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PRINCIPLES OF MECHANICS 



PRINCIPLES 

of 

MECHANICS 



BY 

JOHN L. SYNGE 

Professor of Mathematics 
Carnegie Institute of Technology 

AND 

BYRON A. GRIFFITH 

Assistant Professor of Mathematics 
University of Toronto 



SECOND EDITION 



NEW YORK TORONTO LONDON 

McGRAW-HILL BOOK COMPANY, INC. 

1949 



PRINCIPLES OF MECHANICS 

Copyright, 1942, 1040, by the McGraw-Hill Book Company, Inc. Printed in 
the United States of America. All rights reserved. This book, or parts thereof, 
may not be reproduced in any form without permission of the publishers. 



PRINTED BY THE MAPLE PRESS COMPANY, YORK, PA. 



Le savant doit ordonner; on fait la science avec 
des faits commc uno maison avcc des pierrcs, 
inais une accumulation de faits n'est pas plus 
une science qu'un tas de picrres n'est une 

maison. 

HKNRI POINCARE: 



PREFACE TO THE SECOND EDITION 

This edition differs in no essential way from the first. The 
principal revision occurs in Chap XIII, where the account of 
the motion of a particle in an electromagnetic field has been 
completely rewritten. The treatment of principal axes of inertia 
in Chap XI has been amplified, and some revisions have been 
made in the treatments of Foucault's pendulum, the spinning 
projectile, and the gyrocompass. The emphasis on units and 
dimensions has been increased by the inclusion in the earlier part 
of the book of a few short paragraphs, with references to the 
Appendix, where these matters are discussed in detail. A few 
additional exercises have been inserted, and numerous minor 
corrections have been made. We wish to thank all those readers 
who have contributed to the improvement of this second edition 
by their suggestions, arid, in particular, Professors L. Infeld, 
A. E. Sehild, and A. Weinstein. 

JOHN L. SYNGE 
BYRON A. GRIFFITH 

PITTSBURGH, PA. 
TORONTO, ONT. 
July, 1948 



PREFACE TO THE FIRST EDITION 

In a sense this is a book for the beginner in mechanics, but in 
another sense it is not. From the time we make our first move- 
ments, crude ideas on force, mass, and motion take shape in our 
minds. This body of ideas might be reduced to some order at 
high school (as crude ideas of geometry are reduced to order), 
but that is not the educational practice in North America. 
There is rather an accumulation of miscellaneous facts bearing 
on mechanics, some mathematical and some experimental, until 
a state is reached where the student is in danger of being repelled 
by the subject, as a chaotic jumble which is neither mathematics 
nor physics. 

This book is intended primarily for students at this stage. 
The authors' ambition is to reveal mechanics as an orderly self- 
contained subject. It may not be quite so logically clear as pure 
mathematics, but it stands out as a model of clarity among all 
the theories of deductive science. 

The art of teaching consists largely in isolating difficulties 
and overcoming them one by one, without losing sight of the 
main problem while attending to the details. In mechanics, 
the main problem is the problem of equilibrium or motion under 
given forces the details are such things as the vector notation, 
the kinematics of a rigid body, or the theory of moments of 
inertia. If we rush straight at the main problem, we become 
entangled in the details and have to retrace our steps in order to 
deal with them. If, on the other hand, we decide to settle all 
details first, we are apt to find them uninteresting because we 
do not see their connection with the main problem. A compro- 
mise is necessary, and in this book the compromise consists 
of the division into Plane Mechanics (Part I) and Mechanics in 
Space (Part II). These titles must, however, be regarded only 
as rough indications of the contents. Part I includes some of 
the easier portions of three-dimensional theory, while Part II 
contains an introduction to the special theory of relativity, with 
mechanics in only one spatial dimension ! 



X PREFACE TO THE FIRST EDITION 

There is, of course, nothing novel in regarding plane mechanics 
as the preliminary field; but it is rather unusual to divide the 
subject in this way in a single volume, or even in a sequence of 
volumes. It has made the task of writing more difficult, but 
the authors have felt it worth while. Many of the most interest- 
ing results in statics and dynamics belong to the plane theory, 
and it is unfair to deny the reader access to them until he has 
mastered the more elaborate technique required for three 
dimensions. 

Part I is complete in itself and might be used as a textbook 
in plane statics and dynamics, with some excursions into three- 
dimensional theory. Vector notation is introduced, but used 
sparingly. The reader should have a fair knowledge of calculus, 
elementary differential equations, and some analytical geometry. 
Practical experience in physics is not essential but very desirable; 
mechanics is at root a physical subject and should not be treated 
merely as an excuse for the exercise of mathematical techniques. 
In Part II the language of vectors is used extensively. A 
knowledge of three-dimensional analytical geometry is required 
and greater power in the use of mathematical processes. This 
part is complete in itself, except for occasional references to 
Part I. The selection of particular applications follows con- 
ventional lines, except for one novel feature a section on electron 
optics. Chapters on Lagrange's equations and on the special 
theory of relativity are included. 

The book has developed from lectures delivered by both 
authors to Honor Students in their second and third years at the 
University of Toronto. These lectures cover about 110 periods 
of 50 minutes, and it has been found that the work can be done 
fairly adequately in that time. But this does not allow suffi- 
ciently for the working of problems with the classes; it is felt that 
150 periods might well be spent on the contents of the book, 
were it not for other demands on the students' time. 

Each chapter is followed by a summary. The summaries to 
the chapters dealing with ijiethods are naturally the more funda- 
mental there is little hope of being able to attack problems 
unless one is thoroughly familiar with the general principles 
outlined there. On the other hand, the summaries to the chap- 
ters dealing with applications are intended to provide only a 
synopsis of what has been done. 



PREFACE TO THE FIRST EDITION xi 

Many of the exercises are taken with permission from examina- 
tion papers set in the University of Toronto and printed by 
the University Press. In each set of exercises, the first few 
problems are so simple that failure to solve them will reveal a 
lack of understanding of basic methods, rather than a deficiency 
in skill and ingenuity. 

The equations are numbered in such a way that, when read 
as decimals, they stand in their proper order. The integer 
represents the chapter, the first decimal place represents the 
section, and the last two decimal places the position of the equa- 
tion in the section. 

Debts to other textbooks are too numerous to acknowledge. 
But we would like to pay tribute to two books and recom- 
mend them to the reader who wishes to pursue the subject 
further. They are E. T. Whittaker's Analytical Dynamics 
(Cambridge University Press) and P. AppelPs Mgcanique 
rationnelle (Gauthier-Villars) . These books have suggested the 
possibility of reconciling in a textbook on mechanics two opposing 
goals the reduction of the subject to a compact and classified 
form and its exposition with sufficient fullness to make the 
arguments easy to follow. 

We gratefully acknowledge assistance and advice received from 
our colleagues, Professor H. S. M. Coxeter, Professor A. F. 
Stevenson, Dr. A. Weinstein, and Mr. A. W. Walker. We are 
under a particular debt to Professor L. Infeld, who read most of the 
manuscript and has been unsparing in frank criticism and sug- 
gestions; if we have succeeded in avoiding dullness and obscurity, 
it is due in no small measure to him. 

J. L. SYNGE 
B. A. GRIFFITH 
TORONTO, ONTARIO 
MEDICINE HAT, AIJBEUTA 
December, 1941 



PART I 
PLANE MECHANICS 



CHAPTER I 

FOUNDATIONS OF MECHANICS 
1.1. SOME PHILOSOPHICAL IDEAS 

Why do we study mechanics? There are at least three reasons. 
First, we live in an age of machinery, which cannot be designed 
without a knowledge of mechanics; in fact, it is the most funda- 
mental subject in engineering. Secondly, mechanics plays a 
basic part in physics and astronomy, contributing to our knowl- 
edge of the working of nature. Thirdly, the mathematician is 
interested in mechanics, both in the logic of its foundations and 
in the methods employed; a considerable portion of mathematics 
was developed for the express purpose of solving mechanical 
problems. 

The subject of mechanics is not a mere collection of facts. 
From certain simple hypotheses an elaborate theory is built up. 
Anyone who has studied the subject should be able to answer 
questions of interest to engineers and physicists; that is to say, 
he should be able to apply his knowledge. But he should also 
have a fair idea of the logical structure. A successful textbook 
has to steer a middle course between undue concentration on the 
mere working out of problems on the one hand, and an over- 
elaborate development of logical structure on the other. 

The two ways of thinking. 

What the student of mechanics requires more than anything 
else is the development of a certain point of view which is difficult 
to describe in a few words. Since the reader is expected to have 
a fair knowledge of geometry, it will be helpful to consider the 
ways in wjiich we think about that subject. 

Every student of geometry learns to think in two ways. 
First, there is the physical way, in which a point is a small dot 
on a sheet of paper, a straight line a mark made by drawing 
a sharp pencil along a straight edge, a circle a mark made by a 
pair of compasses, and so on. Secondly, there is the ideal or 

3 



4 PLANE MECHANICS [Sac. 1.1 

mathematical way, in which a point is no longer a dot on paper, 
but an ideal thing which the dot serves only to suggest. Anyone 
who uses geometry has both these ways of thinking at his disposal, 
switching from one to the other without confusion. The engineer 
and the physicist generally think in the physical way, but when 
there is a theorem to be proved they subconsciously switch to the 
mathematical way. On the other hand the mathematician will 
think primarily in the mathematical way, but he will change to 
the physical way when he wants to aid his thought with a 
diagram. 

This duality in point of view is confusing to the beginner 
in geometry. But it is fortunate that he has to face this difficulty 
at an early stage in his career, because it prepares him for a 
similar duality in mechanics, about which he has also to learn 
to think in two different ways. 

First, there is the physical way. We think of actual physical 
things, natural or man-made. We seek to understand the laws 
governing their behavior and to predict how they will behave 
under given circumstances to be able to trace the paths of 
comets in advance, or design machinery and bridges with con- 
fidence as to their behavior when constructed. 

On the other hand, there is the mathematical way. Often 
without realizing it consciously, the physicist, astronomer, or 
engineer slips over from the physical way of thinking to the 
mathematical. Thus the astronomer may treat the earth as a 
perfect sphere an abstract mathematical concept which does 
not exist in nature or the engineer may discuss a wheel as 
if it were a perfect circle. 

The transition from the physical to the mathematical and 
back again is a source of more confusion than may be suspected, 
but it is unavoidable. There is no doubt that the physical 
way of thought is the more natural; but as long as it is the only 
way, progress is slow. Physical things are very complicated 
and hard to think about. Slowly we come to distinguish between 
properties which are essential and properties which are incidental. 
We learn to simplify problems by forgetting the incidental 
properties and concentrating on those which are essential. 

To illustrate, suppose we are interested in the periodic time 
of a bar suspended from one end, oscillating as a pendulum. 
Which properties of the bar is it essential for us to bear in mind, 



SEC. 1.1] FOUNDATIONS OF MECHANICS * 5 

and which may we neglect as incidental? Can we predict the 
periodic time of oscillation without knowing the material of 
which the bar is constructed? Does the form of the cross section 
of the bar matter? Does it make any difference whether the 
bar is supported on a knife-edge or by bearings? The cautious 
well-informed physicist would say that all these things mattered 
and many others. One material yields more than another, the 
form of cross section influences the distribution of material, and 
a change in the mode of suspension may alter the axis about 
which the pendulum oscillates. But if we were as cautious as 
this we should have no science of mechanics. To start on the 
problem, at any rate, we must simplify it ruthlessly. So we 
think of the bar as a rigid mathematical straight line and the 
support as a fixed mathematical point. Now we have a problem 
which is reasonably simple to handle mathematically. Strictly 
speaking, no properties are incidental. Even the color of the 
bar affects the pressure of light on it; a subway train stopping 
five hundred miles away may cause a vibration in the support and 
affect the motion of the bar. Common sense, which is the accu- 
mulated experience of centuries, gives us some guide as to the 
factors which we may neglect. 

Mathematical models. 

Gradually stripping physical things of attributes which are 
unimportant for the question in hand, we arrive at a mathe- 
matical way of thinking about nature. The particular mathe- 
matical model* to be used on a given occasion depends on that 
occasion. Consider the earth, for example. The simplest model 
of the earth is a particle, a mathematical point with mass. This 
model suffices to obtain the earth 's orbit round the sun, but 
obviously will not do for the discussion of tides or lunar 
eclipses. For these phenomena we may think of the earth as a 
rigid sphere, but this model will not serve for the discussion 
of the precession of the equinoxes (for which we require an ellip- 
soidal rigid body) or for the discussion of earthquakes (for which 
we require an elastic sphere). Thus there are many mathe- 

* The reader will of course understand that when we speak of a "model" 
we do not mean an actual physical reproduction on a small scale. We use 
the word for want of a better to describe our simplified mental picture 
of a physical object. 



6 PLANE MECHANICS [SEC. 1.1 

matical models for the earth, and the one which we choose 
depends on the question we are discussing at the moment. 

In fact, mechanics and indeed all theoretical science is a 
game of mathematical make-believe. We say: // the earth 
were a homogeneous rigid ellipsoid acted on by such and such 
forces, how would it behave? Working out the answer to this 
mathematical question, we compare our results with observa- 
tion. If there is agreement, we say that we have chosen a good 
model; if disagreement, then the model or the laws assumed are 
bad. 

Let us now sum up the general procedure in theoretical 
mechanics in the following five steps. 

(1) A physical system is an object of curiosity; we wish to 
predict its behavior under various circumstances. (The system 
in question might be a pendulum, or a pair of stars attracting 
one another.) 

(2) An ideal or mathematical model of the physical system is 
constructed mentally. (The pendulum is regarded as a rigid 
straight line, and the stars are regarded as two particles.) 

(3) Mathematical reasoning is applied to the mathematical 
model. (This means that differential or finite equations are set 
up and solved. Formulas are developed to give answers to 
interesting questions, such as those concerning the periodic 
time of the pendulum or the orbit of one star relative to the 
other.) 

(4) The mathematical results are interpreted physically in 
terms of the physical problem. 

(5) The results are compared with the results of observation, 
if possible. 

Certain remarks should be made about these five steps. First, 
(1) implies a physical curiosity. In spite of the fact that theo- 
retical mechanics is a part of mathematics, we should not forget 
that its roots lie in physics and the actual world around us. 

Secondly, as has been remarked above, the construction of 
a mathematical model (2) at once simple and adequate is by no 
means easy in all cases. However, mechanics is an old subject, 
and there is much accumulated experience to fall back on. The 
concepts of particles, rigid bodies, forces, etc. (all mathematical 
idealizations), have be n designed for this purpose. These will 
be discussed ^ v 



SEC. 1.1] FOUNDATIONS OF MECHANICS 7 

Step (3) belongs largely to pure mathematics, requiring no 
particular knowledge of, or interest in, the physical problem. 
Nevertheless, it is often of the greatest assistance to the mathe- 
matician to bear the physical problem continually in mind; in 
this way, methods of attack may be suggested to him. 

The fourth step in general presents no difficulty, provided 
that we are clear as to the things in nature which correspond to 
the things in our mathematical model. 

The technical details of the fifth step belong to experimental 
physics or observational astronomy, and with them we shall 
not be concerned. But we are interested in the fact that the 
conclusions drawn from a mathematical theory are, or are not, 
physically true, within the limits of accuracy of observation. 

It is necessary to distinguish between mathematical truth 
and physical truth. In developing the theory of mechanics, we 
shall try to make the mathematical arguments fairly complete, so 
that we can have confidence that the conclusions follow logically 
from the hypotheses, i.e., that they are mathematically true. 
We should not undertake this work, however, if we had not 
confidence that our conclusions are also physically true, in the 
sense that they agree with observation. A vast accumulation of 
physical results confirms our confidence. Nevertheless, it would 
be too much to claim that all our conclusions are physically valid. 

Attempts to construct a successful model of an atom on the 
basis of Newtonian mechanics have failed. This failure led 
to the invention of quantum mechanics. We may say in gen- 
eral that Newtonian models of small-scale phenomena have 
not been successful, whereas at the other end of the scale we 
find difficulty also in the large-scale phenomena of astronomy. 
In spite of the many triumphs of Newtonian mechanics in 
dynamical astronomy, there remain a few phenomena which 
are in apparent disagreement with it; the best-known concerns 
the orbit of the planet Mercury. This difficulty was overcome 
when Einstein created the general theory of relativity. 

To explore with any degree of completeness the theories 
referred to above would demand a course of study far wider than 
that covered in this book. The reader may feel disappointed 
that at this stage he cannot reach the forefront of our mechanical 
knowledge. To encourage him, however, it may be pointed out 
that as long as the physical problems concern only apparatus 



8 PLANE MECHANICS [Sac. 1.2 

of an intermediate scale, i.e., neither atomic on the one hand nor 
astronomical on the other, one may have complete confidence 
that no experimental technique can reveal any discrepancy 
between observation and the conclusions drawn from New- 
tonian mechanics. This confidence may even be extended to 
astronomy, because there the relativistic effects are extremely 
minute; the vast body of calculations of dynamical astronomy 
are still safely based on Newtonian mechanics. 

Relativity and quantum mechanics not only enable us to 
obtain results which are physically true they also throw light 
on such basic philosophical ideas as simultaneity and causality. 
Chapter XVI contains an introduction to the special theory of 
relativity. The general theory of relativity and quantum 
mechanics both lie outside the scope of this book. 

1.2. THE INGREDIENTS OF MECHANICS 

In any subject there are words which occur again and again, 
like the words "point/ 7 "line," and "circle" in elementary geom- 
etry. As well as these, technical words, there occur ordinary 
words with the meanings of which we are supposed to be familiar. 
When we start a new subject, we are not expected to know 
what the technical words mean. They are introduced with some 
formality, being in fact given definitions. A definition is itself 
only a set of words and may not mean much; the general idea is 
to explain a new thing in terms of things already familiar. 

We are now to try to create mathematical models of physical 
things. We start with a fair general unprecise knowledge of the 
world around us; the places in our minds reserved for the mathe- 
matical models are supposed to be absolutely blank. If we 
opened these places for the actual world to rush in, we should be 
overwhelmed with confusion. We guard the door and admit 
only a few ingredients of simple mathematical character. 

Particles. 

The first thing we admit is a particle. We have seen tiny 
scraps of matter and it is not difficult for us, with our training 
in geometry, to think of a scrap of matter with no size at all, 
but with a definite position; that is a particle. When we have 
to deal with a physical problem in which a body is very small 
in comparison with distances or lengths involved (for example, 



SBC. 1.2] FOUNDATIONS OF MECHANICS 9 

the earth in comparison with its distance from the sun, or the bob 
of a pendulum in comparison with the string), we may represent 
that body in our mathematical model by a particle. 

Mass. 

Primitive trade was a matter of barter; later, money was intro- 
duced as a standard scale for comparison of values, and equiva- 
lence in value is now expressed by equality of price. This 
exemplifies a process of deep importance in science, namely, 
a concentration on some characteristic (value) of a thing and 
its expression by means of a number (price). A barrel of apples 
is very different from a pair of shoes, but they may be equivalent 
if value is the only characteristic in which we are interested. 
That the price is the same expresses complete equivalence as 
far as our purse is concerned. 

Consider now a great variety of bodies pieces of stone, 
iron, gold, wood, etc. and mechanical experiments ' performed 
on them. As examples, we mention two experiments: 

(i) The body is placed in the pan of a spring balance and the 
reading noted. 

(ii) The body is fired from a gun by means of a definite explo- 
sive charge and passes into a block of wood, the resulting dis- 
placement of which is noted. 

If A and B are two pieces of iron, as nearly identical in shape 
and size as it is possible to make them, they will of course give 
the same results when used in any experiment, performed first 
using A and then repeated using B instead. But it is a remark- 
able fact, resting on long experience, that two bodies A and B 
may differ in material, size, shape, etc., and yet give the same 
result in a great variety of mechanical experiments. We then 
say that they are mechanically equivalent. A piece of wood and a 
piece of gold may be mechanically equivalent, just as a barrel of 
apples and a pair of shoes may be equivalent in value. 

Asjwe assign a price to each article of trade, so we may assign 
a numEer to each piece of matter, equality of these numbers 
implying mechanical equivalence. This number is called mass 
and is usually denoted by m. Following the analogy of money, 
based on a standard substance (gold), it is easy to see how a 
scale of mass is to be constructed. We start with a number of 
identical pieces of some standard material such as platinum, 



10 PLANE MECHANICS [SEC. 1.2 

and we assign to the mass of each the value unity (m = 1). 
When n of these pieces are lumped together, we assign to the 
mass of the lump the value n(m = n). By cutting the pieces, 
we can construct bodies with fractional masses and so obtain 
a set of standard bodies of all possible masses. Then, to assign 
a mass to a body A (not of the standard material), we subject it 
to experiments and find that standard body B to which it is 
mechanically equivalent. We then say that the mass of A 
is the same as the mass of B. 

The comparison of masses is usually made by weighing as 
in the experiment (i) mentioned above, except that for reasons 
of accuracy the spring balance is replaced by a laboratory balance. 
Thus, in practice, two bodies are said to have the same mass 
when they have the same weight. 

The above considerations deal with physical bodies. In the 
mathematical model in which these bodies are represented by 
particles, we are to regard each particle as having attached to it 
a positive number m, its mass, which does not change during the 
history of the particle. 

In dealing with a system of particles, we define the mass of 
the system to be the sum of the masses of the particles which 
compose it. 

Rigid bodies. 

We have now admitted as a mathematical model the particle 
with mass. The next thing to consider is the rigid body. 

It is a matter of common experience that bodies may be soft 
like rubber or hard like steel. Even the hardest body, however, 
changes its size and shape by measurable amounts under the 
action of sufficiently great forces. But just as we idealized the 
small body of our experience into the particle with position 
but no size, so we idealize the hard body of our experience into 
the rigid body, which never undergoes any change jrfsize or_shape. 
The rigid body is now admitted as a mathematical mocfeL 

We pause for a moment to examine critically something written 
just above. We spoke of a body changing its size and shape. 
What does this really mean? Suppose, for example, we have a 
bar of steel with two marks on it. Alongside the bar we lay a 
graduated measuring scale and note the readings on the scale 
opposite the two marks on the bar. Then we pull the ends of 
the bar and note the readings again. The difference between 



SBC. 1.2] FOUNDATIONS OF MECHANICS 11 

them is greater than it was before; hence we say that the bar 
has increased in length. 

However, an argumentative person might assert that this was 
an incorrect statement ; he might hold that the length of the bar 
was the same as before but that the measuring scale had shrunk. 
We cannot say that he is wrong in taking this point of view until 
we clarify our ideas as to the meaning of the word "length." 

The idea of length is one that involves the comparison of 
two bodies. We decide once for all on a unit of length by making 
two marks on a piece of metal and stating conditions with regard 
to temperature and pressure under which measurements are 
to be made with this piece of metal. We were perhaps a little 
hasty in admitting a rigid body as a mathematical model, because 
there is no sense in talking about a single rigid body; we must 
have some means of measuring it and testing that it is rigid. 
So when we admit the rigid body, we shall at the same time admit 
a measuring scale. When we say that a body is rigid, we mean 
that measurements of distances between marks on it always 
have the same values, the measurements being made with the 
measuring scale. 

Events. 

The word event is familiar in ordinary speech. It usually 
denotes something a little out of the ordinary, something that 
occurs in a fairly limited region of space and is of fairly short 
duration. Thus a football game or the arrival of a train might be 
described as an event. The word has now acquired an idealized 
scientific meaning, the idealization involved being rather similar 
to that by which we created the concept of particle. Instead of 
occupying a fairly limited region in space, an event (in our mathe- 
matical model) occurs at a mathematical point; and instead of 
being of fairly short duration, it occurs instantaneously. We do 
not carry over into our mathematical model the slightly dramatic 
meaning attached to the word in ordinary life. Anything that 
happens may be called an event. Even the continued existence 
of a particle forms a series of events. 

Frames of reference. 

In describing an event in ordinary life, it is usual to specify 
the place and time. Thus it is recorded of the sinking of a 
ship that it occurred at a certain latitude and longitude, and 



12 PLANE MECHANICS [SBC. 1.2 

at a certain Greenwich mean time. Latitude and longitude 
define position on the earth's surface; we are here using the earth 
as a frame of reference. This is the most familiar frame of refer- 
ence, but others may be used. Astronomers prefer a frame of 
reference in which the sun is fixed and which does not share in 
the earth's motion of rotation. Also, the interior of a train, 
streetcar, elevator, or airplane may be used. The essential 
thing about a frame of reference is that it should be fairly rigid. 

In our mathematical model, we employ a rigid body as frame 
of reference. As we may introduce any number of rigid bodies 
moving relative to one another, we have thus at our disposal 
any number of frames of reference. Selecting one of these 
and taking rectangular axes of coordinates in it, we assign to 
any event a set of three numbers x, y, z, the coordinates in the 
frame of reference of the point where the event occurs. 

Time. 

An event has not only position; it also has a time of occurrence. 
This we have now to consider. 

The possibility of repeating an experiment forms the basis 
of experimental science. It is assumed that, if an experiment is 
repeated under the same conditions, the same results will be 
obtained. Consider, for example, a tank of water drained 
through a hole in the bottom, and then refilled and drained again. 
Strictly speaking, it is impossible to reproduce conditions 
exactly, and we have to use judgment to decide whether the new 
conditions are sufficiently near the old. But in an ideal sense 
we may think of an experiment repeated over and over again 
under exactly the same conditions. 

To define time, we think of some experiment which can bo 
repeated over and over again, a new experiment starting just 
when the preceding one ends. Denoting time by t, we assign 
the value t = to the beginning of the first experiment, t 1 
to the beginning of the second experiment, t = 2 to the beginning 
of the third experiment, and so on. The repeated experiment 
thus forms a clock for the measurement of time; we shall call the 
unit of time given by some such ideal experiment a Newtonian 
unit. This is the procedure actually adopted in practice. In 
a watch, the experiment is an oscillation of the balance wheel; in 
a pendulum clock, it is an oscillation of the pendulum. In 



SBC. 1.2] FOUNDATIONS OF MECHANICS 15 

As has been pointed out already, we are not to expect a 
mathematical model to have all the complexity of nature. The 
model which we shall use resembles in some ways the modern 
physicist's concept of a solid body, but it is greatly simplified. 
It was invented long before the development of modern atomic 
physics, and was originally supposed to be a more complete 
representation of nature than we now know it to be. Neverthe- 
less, it enables us to predict to a high degree of accuracy an 
immense number of phenomena; it is in fact the basis of a great 
deal of Newtonian mechanics. 

This mathematical model of a solid body is discontinuous 
a collection of a vast number of particles. In a rigid body the 
distances between the particles remain invariable, but in an elastic 
body these distances may change. Since this model involves 
a very large number of things, statistical methods may be used; 
instead of following individual particles, we may direct our 
attention to their average behavior. In fact, we mentally replace 
the discontinuous body, consisting of a great number of par- 
ticles, by a continuous distribution of matter. This simplifies 
the determination of mass centers and moments of inertia, 
because the methods of integral calculus can then be used. 

To avoid lengthy and perhaps uninteresting arguments, we 
shall leave certain gaps in the logical development of our subject. 
We shall not give arguments of a statistical nature in order to 
pass from a result established for a discontinuous system to the 
corresponding result for a continuous one. It is usually easier 
to establish general theorems for discontinuous systems and to 
solve special problems for continuous systems. 

Force. 

Let us now introduce into our mathematical model the concept 
of force, idealizing as usual from our somewhat vague physical 
concepts. Our primitive concept of force arises out of our sensa- 
tion of muscular exertion. We push and pull objects, sometimes 
with small exertion, sometimes with great effort. But the same 
effects as those produced by muscular effort may be produced in 
other ways. In this machine age, direct muscular effort is used 
to a great extent only to control much greater forces due to the 
pressure of steam, the weight of water, the explosive pressure of 
gasoline, or forces of electromagnetic type. One also admits the 



16 PLANE MECHANICS IBC. i.z 

existence of huge forces beyond human control, such as the gravi- 
tational attraction exerted by the sun on the earth. 

On the basis of our experience with simple muscular forces, 
we think of the idealized force of our mathematical model as 
something which has 

(i) a point of application, 

(ii) a direction, 

(iii) a magnitude. 

For the development of general results in theoretical* mechanics, 
it would be sufficient to represent the magnitude of a force by a 
letter, standing for some unspecified numerical value. But when 
we wish to make predictions regarding a physical system subject 
to forces, we require a definite procedure by means of which we 
may assign numerical values to their magnitudes. We must, 
in fact, define a unit of force. 

There has been some controversy about this question. Though 
all are agreed as to the form of theory which we should ultimately 
obtain, there has been disagreement regarding the proper order 
of introduction of the various parts of the theory. Thus we 
might assume a statement A as an axiom and deduce a statement 
B from it, or alternately we might assume B and deduce A. 
The order of presentation chosen in this book seems to the 
authors the most natural; but it is hoped that the reader will 
explore for himself the possibility of a different approach. * 

We define the unit of force in terms of a stretched spring; 
it is that force which produces some standard extension in some 
standard spring. Later we shall link up the unit of force with 
the units of mass, length, and time; but for the present the unit 
is to be regarded as arbitrary. 

To measure a force, we examine the extension which it produces 
in a battery of standard springs side by side, all identical with 
one another. If the standard extension is produced in n springs, 
then the force is of magnitude n. If the magnitude of the force 
in question is not an integer, we reproduce it m times so as to get 

* See treatments in E. T. Whittaker, Analytical Dynamics (Cambridge 
University Press, 1927), p. 29; H. Lamb, Statics (Cambridge University 
Press, 1928), p. 12, and Dynamics (Cambridge University Press, 1929), p. 17; 
J. S. Ames and F. D. Murnaghan, Theoretical Mechanics (Ginn and Com- 
pany, Boston, 1929), p. 104. For a critical and historical account of the 
development of mechanics, see E. Mach, The Science of Mechanics (Open 
Court Publishing Company, Chicago, 1919). 



SEC. 1.3] FOUNDATIONS OF MECHANICS 17 

the standard extension in some number n of standard springs; 
then the magnitude of the force is n/m. 

Having thus given a means of measuring force, we may con- 
struct a simplified apparatus. Taking any spring, fixed at 
one end, we mark its extensions under the action of measured 
forces. In this way, we calibrate the spring; the calibrated spring 
may be used directly for the measurement of a force. 

We have preferred to use the tension in a spring rather than 
the weight of a body as the foundation of our definition of force. 
It is customary for many practical purposes to speak of a force 
of so many pounds weight (Ib. wt.); by a force of 10 Ib. wt., 
we mean a force equal to the weight of a mass of 10 Ib. Although 
this* practice is convenient and adequate for many purposes 1 ; 
it is open^to objection on the following ground: If the weight 
of a body is measured by means of a calibrated spring at two 
different latitudes, the results are not the same (see Sec. 5.3). 
A definition of force based on the extension of a spring gives a 
consistent theory without contradictions, whereas a definition 
based on weight would involve us in explanations as to why a 
spring, showing the same definite extension in Toronto and in 
Panama, should exert different forces in the two places. 

In describing particles, rigid bodies, and forces we have intro- 
duced the basic ingredients of mechanics. As we proceed, other 
ingredients will appear, but it is remarkable how much of the 
subject turns on the simple concepts just mentioned. 

1.3. INTRODUCTION TO VECTORS. 
VELOCITY AND ACCELERATION 

Definition of a vector. 

In order to reproduce a game of chess, we must be able to 
describe the moves. There is, of course, an accepted way of 
doing this, but we shall describe another. Let letters A, B, 
C, (supplemented with other symbols to make up 64) 
be assigned to the squares of the board, one letter to each square. 

Then symbols such as AB, CF, UB will represent definite moves, 

> 
the symbol AB, for example, meaning that a piece is moved from 

the square A to the square B. 

More generally, if we carry a particle from a position A to a 
position B in space, the operation which we perform may be 



18 PLANE MECHANICS [Sac. 1 3 

> 

represented symbolically by AB. The directed segment drawn 
from A to J3, or the carrying operation or indeed any physical 
quantity which can be represented by the directed segment is 

called a vector,* and the symbol A B is used for any of them. 

A vector AB has the following characteristics: 

(i) an origin or point of application (A); 

(ii) a direction (defined analytically by the three direction 

cosines of AB with respect to rectangular axes); 

(iii) a magnitude (the length AB). 

A number of fundamental physical quantities have these char- 
acteristics for example, a force or the velocity of a particle; 
each of these quantities may be represented by a directed seg- 
ment and is therefore a vector. They are to be distinguished 
from quantities such as mass or kinetic energy, which do not 
involve the idea of direction and are described each by a single 
number. Quantities of this latter type are called scalars. 

It is convenient to employ the word "vector" in a slightly 
wider sense than that given above and to define the following: 

(i) free vector; 

(ii) sliding vector; 

(iii) bound vector. 

A. free vector is any one of a system of directed segments having 
a common direction and magnitude but different origins. A 
physical quantity equally well represented by any one of such 
directed segments is also called a free vector. Such, for example, 
is the displacement, without rotation, of a rigid body, which 
is equally well represented by any one of the directed segments 
giving the displacements of its various points. 

A sliding vector is any one of a system of directed segments 
obtained by sliding a directed segment along its line. A physical 
quantity equally well represented by any one of such directed 
segments is also called a sliding vector. Such, for example, 
is a force acting on a rigid body, which (by the principle of the 
transmissibility of force proved on page 64) may equally well 
be applied at any point on its line of action. 

A bound vector is a unique directed segment, or a physical 
quantity so represented. Such, for example, is a force acting on 

* The word "vector" is derived from the Latin veho, I carry. 



SEC. 1.3] FOUNDATIONS OF MECHANICS 19 

an elastic body; we cannot in general alter this force, by any 
displacement of the directed segment representing it, without 
changing its effect. 

By the word " vector," without an adjective, we shall generally 
understand "free vector"; but where we have to speak of bound 
or sliding vectors, it will be unnecessary to use the qualify- 
ing adjective when it may be understood from the context. 
Throughout the rest of this section, the vectors are to be regarded 
as free. 

Notation. 

A vector is indicated in print by a boldface letter (P): 
in manuscript work the symbol may be underlined (P) or an 

arrow may be written on top (P). Its magnitude is denoted by 
the same letter in ordinary type, or by an unmarked symbol 
in manuscript work (P). A vector of unit magnitude is called 
a unit vector. To refer to a bound or sliding vector, we may 
write "P acting at the point A 11 or "P acting on the line I/," 
if there is any doubt as to the origin or line. 

Two vectors are equal to one another when they may be 
represented by equal parallel directed segments with the same 
sense. We use the usual sign of equality and write 

P = Q. 

The sign of equality carries the usual algebraic property: vectors 
equal to the same vector are equal to one another. 

Multiplication of a vector by a scalar. 

Let P be a vector and m a scalar. We define the product of 
ra and P (written raP or Pw) as follows: If m is positive, then 
rriP has the same direction as P and a magnitude wP; if m is 
negative, then mP has a direction opposite to that of P and a 
magnitude raP. 

We write (-l)P = P; thus P is the vector P reversed. 

Addition of vectors^ 

The sum of two vectors P and Q is written P + Q; it is defined 


as the vector represented by the diagonal AD of a parallelogram 

of which two adjacent sides AB, AC represent P and Q, respec- 



20 



PLANE MECHANICS 



[SBC. 1.3 



tively (Fig. 1). Obviously, an alternative way of constructing 
P + Q is the following (Fig. 2) : Draw a segment AB to represent 

P, and from its extremity draw BD to represent Q; then A D 
represents P + Q. 

This is a mathematical definition of P + Q. It' does not 
contain the implication that P + Q is the physical resultant of 
P and Q, although in almost all cases we shall find that P + Q 
is actually the physical resultant. Finite rotations are the out- 
standing exceptions; a finite rotation is a vector, but the resultant 
of two finite rotations is not the sum of the vectors (cf. Sec. 10.5). 





FIG. 1. Addition of vectors by 
parallelogram. 



FIG. 2. Addition of vec- 
tors by triangle. 



When two vectors P and Q have the same direction or opposite 
directions, the parallelogram constructed to give their sum 
collapses into a straight line. But that does not prevent us 
from applying the above definition, regarded as a limiting process. 
It is easily seen that, if P and Q have the same direction, then 
P + Q has also that direction and a magnitude P + Q; if they 
have opposite directions and P is the greater, then P + Q has 
the direction of P and a magnitude P Q. Comparing this 
with the definition of the product of a vector by a scalar, we 
find in particular that 

P + P = 2P, 

and we verify generally that the multiplication of a vector by a 
scalar is distributive both with respect to the* scalar and to the 
vector; this means that we have 



(1.301) 
(1.302) 



(m + ri)P = raP + wP, 
m(P + Q) = mP + mQ. 



SEC. 1.3] FOUNDATIONS OF MECHANICS 21 

It is an immediate consequence of the definition that the 
addition of vectors is commutative, that is, 

P + Q = Q + P 

The subtraction of vectors is immediately effected by writing 
P - Q = P + (-Q) 

and applying the rule for addition. The difference between P 
and Q is easily constructed as follows : p 

Draw AB, AC to represent P, Q, re- 




spectively (Fig. 3); then CB repre- Q 

sents P - Q. C 

Applying the rule for subtraction FlQ - 3.-subtaotion of vectors, 
to the case P P, we obtain a vector of zero magnitude, which we 
denote by 0, so that 

P - P = 0. 

We call the zero vector; all vectors of zero magnitude are regarded 
as equal to one another. 

Any unfamiliar symbol containing vectors must be approached 
with caution. .It may mean nothing at all (for example, we never 
attempt to define the sum of a scalar and A 
a vector, m + P) ; on the other hand, it 
may be given a meaning. On the basis 
of previous definitions, P + Q + R has 
no meaning, because we have defined the 
sum of two vectors, not three. But 

(P + Q) + R 

has a meaning, if we regard the paren- 
theses as carrying the instruction to add 
P and Q, and then add R to that sum; ^ 

FIG. 4. The associative 
P ~t~ (Q "4" R) property of vector addition. 

has a meaning, also. It is then easy to see that 

(P + Q) + R = P + (Q + R), 

> > 
by means of the construction shown in Fig. 4, where AB, BC, CD 

represent P, Q, R, respectively, and AD either of the above sums. 




22 PLANE MECHANICS [SBC. 1.3 



Thus a unique meaning can be attached to P + Q + R. We say 
that vector addition is associative. 

Exercise. If P 4- 2Q = R, 

P - 3Q - 2R, 

show that P has the same direction as R, and Q the opposite direction. 

Components of a vector. 

Let L be a straight line and P a vector represented in Fig. 5 
by AB. If we draw through A and B planes perpendicular to L, 

these planes cut off on L a directed segment CZ), the orthogonal 


projection of AB] CD is a common perpendicular to the pair of 

planes. If we take a different directed segment A B f to represent 

> * 

P, we get a projection C'D' on L. Now if we give AB and the 

B' 
B 

^ 
P 




C D C 7 T)' 

FIG. 6. The component of a vector on a lino. 

pair of planes associated with it a displacement (without rotation) 
which carries A to A', then B will go to B' and the pair of planes 

associated with A B will coincide with the pair of planes asso- 
ciated with A'B'. CD will become a common perpendicular 

to the latter pair of planes, and so CD and C'D' are equal in 
magnitude and direction. Thus the vector obtained by project- 
ing on L a directed segment representing P is independent of the 
particular segment chosen. Writing Q for the vector repre- 
sented by CD or C'D', we say that Q is the vector component of P 
on!/. 

Of the two senses on the line L, let us pick out one and call 
it the positive sense, the other being negative; the line L is then 
said to be directed. Let i be a vector of unit magnitude, lying 
on L and pointing in the positive sense. Then it is possible to 



SEC. 1.3] FOUNDATIONS OF MECHANICS 

express the projection Q in the form 



23 



where c is a scalar, positive if Q has the same sense as i and 
negative if Q has the opposite sense to i. The scalar c is called 
the scalar component of P on the directed line L. 

Since Q is represented by any common perpendicular (in the 
proper sense) to the projecting planes, it is easy to see that the 
scalar component of P on L is P cos 0, where is the angle 
between P and the positive sense of L. 

In speaking of a component (without qualification), it will be 
clear from the context whether the vector or scalar component 
is to be understood. 

Unit coordinate vectors. 

In a frame of reference S, let Oxyz bo rectangular Cartesian 
axes. Let i, j, k be unit vectors lying on Ox, Oy, Oz, respectively, 
each in the positive sense. The sot i, j, k is called a unit ortho- 
gonal triad. Let P be any vector, and PI, P 2 , P 3 its components 
on Ox, Oy, Oz, respectively. These 
components, obtained by projec- 
tion, are independent of the par- 
ticular directed segment chosen to 
represent P; we may therefore take 
the representative segment with its 
origin at (Fig. 0). From the rule 
of vector addition, it is clear that 

(1.303) P = Pii + P 2 j + 



P 3 k 



'P.I 



FIG. C. -Resolution along unit 
coordinate vectors. 



This reduction of a vector to the 

sum of three vector components 

along a unit orthogonal triad is of 

great service in mechanics, for it 

represents the link between the vector methods and the more 

usual methods of analysis. Vector notation is only a shorthand 

for the expression of fairly general statements. In the end, we 

must work in ordinary numbers, and the above formula is the 

bridge by which we pass from vectors to ordinary numbers. 

We note that the magnitude of a vector P is expressed in terms 
of its scalar components by 



24 PLANE MECHANICS [SEC. 1.3 



(1.304) P = VPl + P\ + PI 

The components are given in terms of P and X, /z, *>, the direction 
cosines of P, by 

(1.305) P! = AP, P 2 = /*P, P 3 = *P. 

The following facts are important but may be left to the reader 
to verify, using Fig. 2 in the case of the second: 

(i) The components of mP are mPi, raP 2 , mP 3 . 

(ii) The components of P + Q are Pi + Qi, P 2 + Q 2 , PS + Q 3 . 

If PI, P 2 , Pa are the components of a vector P on the coordinate 
axes and L is a directed line with direction cosines A, /u, v, then 
the angle between P and L is given by 



Hence the scalar component of P on L is 

(1.306) P cos - APi + /*P 2 + pPs. 

Exercise. The components of a vector on axes Oxy in a plane are X, Y. 
What are the components X', Y r on axes Ox'ij', obtained by rotating Oxy 
through an angle 0? 

Position vector. 

Let A be a particle, moving relative to a frame of reference S in 

which Oxyz are rectangular Cartesian axes. The vector OA is 
called the position vector of A relative to 0. If we denote it by r 
and the coordinates of A by x, y, z, we have 

(1.307) r = xi + yj + zk, 

where i, j, k is the unit orthogonal triad along the axes. 

As the particle moves, the vector r changes. In fact, r may be 
regarded as a vector function of the time t\ we may express this 
by writing 

r = r). 

Differentiation of a vector. 

The above considerations lead naturally to a more general 
concept, namely, a vector P which is a function of a scalar u, a 
relation expressed by writing 

P = P(u). 




SEC. 1.3J FOUNDATIONS OF MECHANICS 25 

We need no longer think of P as a position vector or of u as the 
time. 

We are familiar with the idea of differentiating a scalar function 
of a scalar: can we enlarge the familiar method to obtain a process 
for differentiating a vector function 
of a scalar? 

We may follow the familiar plan 
almost word for word. We consider 
two values of the parameter, u and 
u + Aw, and the corresponding in- _^^^ 

crement in P, P* *^/ AP 

FIG. 7. Differentiation of a 
AP = P(u + Aw) ~ P(u). vector. 

We multiply by I/Aw, to form the quotient AP/Aw (Fig. 7) and 
let Aw tend to zero. Thus we get a limiting vector 

(1.308) ^ = lim ~ t 

V ' du Att-K) AW 

which we call the derivative of P with respect to w. 

To find the components of the derivative we introduce unit 
coordinate vectors, so that 

(1.309) P = P 1 i+P 2 j+P 3 k; 

Pi, P 2 , PS are scalar functions of w. Increasing w to w + Aw, we 
have 

P + AP = (Pi + APj)i + (P 2 + AP 2 )j + (P 3 + AP,)k. 
Subtraction gives 

AP = APii + AP 2 j + AP 3 k. 
Dividing by Aw, we get 

^^^;+^J?i4-^L 3 k 
Aw Aw "*" Aw J ^ Aw ' 

and so, letting Aw tend to zero, 



the derivatives on the right being of course derivatives of scalars 
the usual derivatives of the differential calculus. Thus the 



26 PLANE MECHANICS [SEC. 1.3 

components of dP/du are 

dl\ f dP*^ dP* 
du' du 1 du 

In words, the components of the derivative are equal to the 
derivatives of the components. 

The following results are easy to prove, either directly from 
the definition of the derivative or from (1.310): 



-7- p = -T- P ~T> 

du r du du 

where P, Q are vector functions of u, and p is a scalar function 
of u. 

It will be observed that (1.310) can be obtained directly from 
(1.309) by differentiation, using the rules (1.311); in this process 
the vectors i, j, k are treated as constants. 

To avoid possible confusion, let us ask tho question: What do 
we mean by saying that a vector Q is constant? This is meaning- 
less without a statement (explicit or understood) regarding the 
frame of reference employed. In a frame of reference S, a vector 
Q is constant if it may be represented permanently by a directed 
segment joining two points fixed in S. But viewed from another 
frame of reference this same vector Q may not be constant. 
Thus, in (1.309), i, j, k are constants in S because they are unit 
vectors along the axes; they may not be constants in another 
frame of reference moving relative to S. 

If the vector P(u) is of constant magnitude, then the triangle 
shown in Fig. 7 is isosceles. As AM tends to zero, the vector 
AP/Aw tends to perpendicularity with P, or, in other words, for 
a vector P(u) of constant magnitude, dP/du is perpendicular to 
P(i). 

Velocity and acceleration. 

Let S be a frame of reference and O a point fixed in S. Let A 
be a moving particle, its position vector relative to O being r. 
We define the velocity of A relative to S to be the vector 

(1.312) q = * 

Using dots to indicate differentiation with respect to time, we have 

(1.313) q = f = *i + j + *k, 



SBC. 1.3] FOUNDATIONS OF MECHANICS 27 

where x, y, z are the coordinates of A relative to rectangular axes 
Oxyz coincident with the unit orthogonal triad i, j, k. Thus the 
components of velocity are (x, y, z). The magnitude q of the 
velocity is called speed.* 
We define the acceleration of A relative to S to be the vector 

(1.314) f - , 
or 

(1.315) f - q = *i + 7/j + zk. 

The components of acceleration are (z, y, z). 
If the velocity is resolved into components, 

q = u\ + vj + wk, 
then 

(1.316) f = u\ + vj + wk. 

We could, of course, continue this process, defining a super- 
acceleration df/dt. However, acceleration is the important 
vector in Newtonian mechanics, and so we stop our definitions 
here. 

As a simple illustration of these ideas, consider a car traveling 
along a straight road. It follows from (1.312) and the definition 
of the derivative of a vector that the velocity q lies along the road 
and points in tho direction in which the car is going. Similarly, 
it follows from (1.311) that the acceleration f lies along the road, 
but now there is an important difference. The vector of accelera- 
tion does not necessarily point in the direction of motion; this 
is the case only if the speed is increasing. If the speed is decreas- 
ing under application of the brakes, the vector f points backward. 
It is not hard to show that when the car rounds a curve the 
velocity continues to point along the road, but the acceleration 
points off the road toward the inside of the curve. 

The formulas (1.313) and (1.315) are particularly useful for 
direct calculation when the motion is described by giving x\ y, z 
as functions of t. Suppose, for example, that 

* Thus velocity is a vector and speed is a scalar. However, when no 
confusion is likely to arise, the word "velocity" is often used in the scalar 
sense to denote the magnitude of the velocity vector; for example, tho 
expression "velocity of light" is used instead of the correct expression 
"speed of light." 



28 PLANE MECHANICS [SEC. 1.3 

x = a cos a>t, y a sin o>2, z = 0, 
where a and o> are constants. Then 

r = a cos ut - i + a sin a? j, 

q = aco sin atf i + aco cos co j, 

f = aw 2 cos ut i aw 2 sin a> j. 

It is easy to see that this is motion in a circle, the velocity pointing 
along the tangent and the acceleration in along the radius. 

Units of velocity and acceleration. 

In the c.g.s. system of units, the components of r are measured 
in centimeters. Any component of the velocity q is obtained by 
dividing a number of centimeters by a number of seconds, and 
the result is expressed as so many centimeters per second or, 
briefly, cm. sec."" 1 The unit of velocity is one centimeter per second. 

Any component of the acceleration f is obtained by dividing a 
velocity component by time or, more precisely, a number of 
centimeters per second by a number of seconds, and the result is 
expressed as so many centimeters per second per second or, 
briefly, cm. sec.~ 2 The unit of acceleration is one centimeter per 
second per second. 

If f.p.s. units are used, the above italicized statements are 
modified by changing the word "centimeter" to "foot." 

Although the c.g.s. and f.p.s. units are commonly used in 
scientific work, there is no necessity to limit ourselves to them. 
The units of length and time may be chosen arbitrarily. Indeed, 
velocities are frequently expressed in miles per hour (m.p.h.). 

Since velocity is obtained by dividing length by time, we say 
it has the "dimensions" [LT~ 1 ]', similarly, acceleration has the 
dimensions [L T~ 2 ]. This notation is discussed in the Appendix. 

Gradient vector. 

When to each point of space, or to each point of a plane, a 
scalar is assigned, we say that we arc dealing with a scalar field. 
Thus the distribution of pressure (or temperature) in the atmos- 
phere at a certain time gives a scalar field. Or consider a map on 
which heights are marked; the height gives a scalar field over 
the map. 

When to each point of space or to each point of a plane a 
vector is assigned, we have a vector field. The wind velocity in the 
atmosphere gives a vector field. 

We shall now show that a scalar field defines an associated 




SEC. 1.3] FOUNDATIONS OF MECHANICS 29 

vector field in a very simple way. For generality, we shall make 
the argument three-dimensional, but the reader will find it 
interesting to consider by way of illustration the map marked 
with heights. 

Let Oxyz be rectangular Cartesian axes and V(x, y, z) a scalar 
field. (For the map, we suppress 
z] V (x, y) is the height of the land 
at the point x, y.) The surfaces 
V = constant are called level sur- 
faces. (On the map the level sur- 
faces become the contour lines.) 
Let A be any point and 8 the level 
surface passing through A (Fig. 8). 
Let us draw the normal to 8 on the 
side on which V increases, and 

t t ., j. . FIG. 8. The gradient vector. 

proceed an arbitrary distance n 

along this normal. Then V is a function of n, and 



at Ay since V is increasing. 

> 
We now introduce a vector AB, called the gradient of V at A, 

or briefly grad V. (It is also denoted by W.) The defining 
properties of grad V are as follows: 

(i) Its direction is along the normal to S at A, in the sense of 
V increasing. 

(ii) Its magnitude is dV/dn, calculated at A. 

In the case of the map, grad V is perpendicular to the contour 
line and points in the uphill sense; its magnitude is the rate of 
increase of height. 

We now proceed to find the components of grad V on the axes 
of coordinates. Let a, ft 7 be the direction cosines of the normal 
to S in the sense of V increasing. Then any infinitesimal dis- 
placement lying in $, i.e., making 

JTr dV , . dV , , dV . A 

^-te^a^****- ' 

also makes a dx + ft dy + y dz = 0. Hence, 

(U17) _, -*-. 

where 4> is some factor of proportionality. As we proceed along 



30 PLANE MECHANICS [Sac. 1.3 

the normal to S at A , we have, since V is a function of x } y, z, 
which in turn are functions of n, 



^ 

dn dx dn dy dn dz dn 
dV tdV-.dV 

"te a+ a^+aT- 

Substitution from (1.317) gives 

/iv 

(1.318) g = <!>(<** + (!* + r 2 ) = *. 

When we substitute this value of <#> in (1.317), we get 

,,o, m W dv dv ' a dv dv dv 

(1.319) ^=5- a7-*5T ar-*ST' 

and so by (1.305) the components of grad V on the axes are 

dV dV dF 
dx 1 dy' dz' 

If i, j, k is the unit orthogonal triad along the axes, then 



(1.320) 
By (1.306) the component of grad V on a directed line L is 



where X, M> ? are the direction cosines of L. Since 

dx dy dz 

\ = -T-> /* = j-> v ^ ~r> 

ds ds ds 

where ds is an element of L, this component is 



+ <>Y.dy , dVdz = dV f 
dx ds dy ds dz ds ds 

The component of grad V in any direction is the rate of change of V 
in that direction. 

We have seen how to obtain a vector field (grad V) from a 
scalar field (7). It is not in general possible to express an 
arbitrary vector field as -the gradient of a scalar field, but we find 
in mathematical physics many vector fields that can be so 
expressed. In Sec. 2.4 we shall discuss fields of force; in most 
cases of physical interest a field of force is the gradient of a scalar 



SBC. 1.4] FOUNDATIONS OF MECHANICS 31 

field of potential energy (with a change of sign). Since the 
description of a vector field requires three functions and a scalar 
field only one function, a considerable simplification results from 
the use of the scalar field. 

Exercise. If in a plane V x 2 + j/ 2 , find the component of grad V at 
the point (1, 0) in a direction making an angle of 45 with the s-axis. 

1.4 FUNDAMENTAL LAWS OF NEWTONIAN MECHANICS 

Let us suppose that we are conducting experiments in which 
small bodies of measured masses are acted on by measured forces. 
We choose some frame of reference and observe the motions of the 
bodies relative to it. We shall use the following notation: 

P = force, 
m = mass, 
f = acceleration. 

The units of force, mass, length, and time are chosen arbitrarily. 

We ask this question: As a physical fact, is there any simple 
relation connecting P, m, and f for the motion of a body? The 
answer to this question is, in general: There is no simple relation. 

We have thought of the experiments as conducted in any frame 
of reference it might be the cabin of an airplane looping the 
loop, or it might be an ordinary laboratory. We ask a second 
question: Is it possible to choose a frame of reference so that 
there is a simple relation connecting P, m, and f ? The answer 
is: Yes. 

One frame of reference yielding a simple relation among 
P, w, and f is the astronomical frame of reference, in which the 
sun* is fixed and which is without rotation relative to the fixed 
stars as a whole. The simple relation is 

P = kmf, 

where A- is a universal positive constant, the value of which 
depends only on the units employed, f 

* More accurately, the mass center of the solar system (see Sec. 3.1); 
actually this point is not far from the center of the sun. 

f This relation is in excellent agreement with astronomical observations, 
but there are exceptions; the orbit of the planet Mercury reveals a minute 
discrepancy. Although few modern astronomers accept this relation as 
absolute physical truth, its validity is so high that it is taken as the basis of 
celestial mechanics. To take a deeper point of view, we must recast our 
whole mode, of thought and follow the general theory of relativity. 



32 PLANE MECHANICS [SBC. 1.4 

The three laws. 

We shall now state the fundamental laws on which Newtonian 
mechanics is based. These are the laws according to which our 
mathematical model of nature works. The laws as stated here 
are equivalent to those used by Newton, but they are expressed 
in a different form.* 

LAW OF MOTION. Relative to a basic frame of reference a particle 
of mass m, subject to a force P, moves in accordance with the equation 

(1.401) P = knd, 

where f is the acceleration of the particle and k a universal positive 
constant, the value of which depends only on the choice of units of 
force, mass, length, and time. 

Any frame of reference relative to which (1.401) holds is 
called Newtonian. 

If P = 0, then f = by (1.401). Since f = dq/dt as in (1.314), 
it follows that q is a constant vector. In words, a particle under 
the influence of no force travels with constant velocity; i.e., it travels 
in a straight line with constant speed. This important special 
case was stated separately by Newton as his first law of motion, 
his second law dealing with the case where P is not zero. Thus, 
his first two laws are included in our law of motion as stated 
above. 

LAW OF ACTION AND REACTION. When two particles exert 
forces on one another, these forces are equal in magnitude and 
opposite in sense and act along the line joining the particles. 

This is often summed up by saying: Action and reaction are 
equal and opposite. 

LAW OF THE PARALLELOGRAM OF FORCES. When two forces P 
and Q act on a particle, they are together equivalent to a single force 
P + Q, the vector sum being defined by the parallelogram construc- 
tion as in Sec. 1.3. 

It is usual to call P + Q the resultant of P and Q; P and Q are 
called the vector components of P + Q. t 

In setting up a system of laws or axioms, it is generally con- 

* For the original form of Newton's laws, see Sir Isaac Newton's Mathe- 
matical Principles, translation revised by F. Cajori (Cambridge University 
Press, 1934), pp. 13, 644. 

f In vSec. 1.3 we used the expression vector component only in the case where 
the components were perpendicular to one another. It is convenient to 
use it also in the present more general sense. 



SEC. 1.4] FOUNDATIONS OF MECHANICS 33 

sidered desirable to make them independent, in the sense that 
no one of them can be deduced from the others. Many attempts 
have been made to deduce the parallelogram of forces, but all 
these deductions require the statement of other laws, which are 
individually simpler than the parallelogram law but rather long 
to state; so, for brevity, we accept the parallelogram law directly. * 

The three laws stated above form the logical basis of mechanics. 
Their full meaning can be understood only by applying them, and 
we shall not delay the development of the subject by further 
general discussion. We should, however, point out their 
significance for the prediction of the results of ordinary laboratory 
experiments. 

Let us suppose that we are discussing the motion of a billiard 
ball which rolls down an inclined plane in a laboratory. Our 
problem is to predict the behavior of the ball by mathematical 
reasoning. On the one hand, we have the actual physical 
apparatus, on the other our mathematical model, in which the 
ball, the plane, and the forces acting are replaced by their 
mathematical idealizations. There is a precise correspondence 
between the physical things and the ingredients of our mathe- 
matical model. 

But here there enters an important question: What physical 
frame of reference corresponds to the basic Newtonian frame 
mentioned in the law of motion the frame relative to which 
(1.401) holds? We have already indicated the answer: The 
physical frame in question is the astronomical frame of reference. 

But we do not want to know the behavior of the billiard ball 
relative to the astronomical frame of reference; we want to know 
its behavior relative to the walls and floor of the laboratory, i.e., 
relative to the earth's surface. Is it legitimate to regard the 
earth's surface as a physical frame corresponding to the New- 
tonian frame of (1.401)? Strictly speaking, it is not. Very 
refined experiments would enable us to detect differences between 
the physical behavior of the billiard ball and the theoretical 
predictions based on that correspondence. These differences 
are due to the rotation of the earth, f But they are very minute 

* For an interesting "proof" of the parallelogram of forces, see W. R. 
Hamilton, Mathematical Papers (Cambridge University Press, 1940), 
Vol. II, p. 284. 

f In Sec. 13.5 we shall discuss some of the dynamical consequences of 
the earth's rotation. 



34 PLANE MECHANICS [SEC. 1.4 

in the vast majority of experiments made in a laboratory and in 
all problems connected with engineering structures; excellent 
predictions may be made from our three laws by taking the 
earth's surface to be the physical frame of reference corresponding 
to the Newtonian frame. 

Units and dimensions. 

In the equation (1.401), a constant k appears, and if the equa- 
tion is left in this form, this constant k will occur throughout our 
dynamical equations. To avoid this, we make k = 1 by a 
special choice of the unit of force; the unit so chosen is called the 
dynamical unit. When it is used, (1.401) reads 

(1.402) P = mi. 

Clearly the dynamical unit of force prodiices unit acceleration 
in unit mass, since if P = 1 and m = 1, then / = 1. In the 
c.g.s. and f.p.s. systems, the dynamical units of force are called 
the dyne and the poundal, respectively. A force of one dyne 
produces an acceleration of one centimeter per second per 
second in a mass of one gram, and a force of one poundal produces 
an acceleration of one foot per second per second in a mass of 
one pound. 

Since, in (1.402), force is the product of a mass by an accelera- 
tion, the dyne may be described as one gram centimeter per 
second per second, or briefly, 

1 dyne = 1 gm. cm. sec." 2 
Similarly, 

1 poundal = 1 Ib. ft. see." 2 

Force measured in dynamical units has the dimensions [MLT~ 2 ]. 

In dynamical problems, we shall always assume that the 
dynamical unit of force is used, so that the law of motion is 
(1.402). In statical problems, on the other hand, we shall leave 
the unit of force arbitrary, since there is no additional simplicity 
to be gained by restricting it. 

At this point the reader is advised to study the Appendix at 
the end of the book in order that he may understand the refer- 
ences in the text to units and dimensions. In particular, atten- 
tion is directed to the use of the theory of dimensions as an easy 
and rapid check against slips in calculation a check which is of 
great value both in elementary and advanced work. In any 
equation in mechanics, all terms must have the same dimensions. 



SEC. 1.5] FOUNDATIONS OF MECHANICS 35 

For example, if we had carelessly derived the formula (cf. 
page 28) 

f = aco cos co i aa> 3 sin &t j, 

a glance would show that something is wrong with this formula. 
The acceleration f has dimensions [LT~ 2 ], aw has dimensions 
[LT~ l ] y aw 3 has dimensions [L7 7 ~ 3 ], while the trigonometrical 
functions and the vectors i ; j are dimensionless. 

1.6. SUMMARY OF THE FOUNDATIONS OF MECHANICS 

The purposes of this first chapter have been (i) to dig down 
to the fundamental physical ideas and (ii) to lay the foundations 
of a logical structure. Before proceeding to the next chapter, we 
now extract and emphasize those concepts and laws which will be 
required later. 

I. The ingredients of mechanics. 

(a) A particle has position and mass (ra). 

(6) A rigid body is a system of particles, the distances between 
which remain unchanged. It may also be regarded as a con- 
tinuous distribution of matter. 

(c) A frame of reference is a rigid body in which axes of coordi- 
nates are taken. 

(d) A force has point of application, direction, and magnitude. 

(e) The units of mass,. length, and time are arbitrary. So also 
is the unit of force, but it is convenient in dynamics to connect 
the unit of force with the units of mass, length, and time. 

II. Vectors. 

(a) Addition of vectors is carried out by means of a parallelo- 
gram. 

(6) The usual simple algebraic rules apply to vectors, but we 
do not yet define the multiplication of vectors by one another. 

(c) A vector function of a scalar may be differentiated; the 
derivative is another vector function. 



P = Pa + P 2 j + 

then Pi, P2, PS are the scalar components of the vector on the unit 
orthogonal triad (i, j, k). 

(e) The components of grad V are 

dV dV } dV m 
dx* dy' dz' 



36 PLANE MECHANICS [Kx. I 

(/) The component of grad V in any direction is dV/ds. 

III. Velocity and acceleration of a particle. 

(a) Position vector: r = xi + yj + zk. 

(b) Velocity vector: q = f = xi + yj + zk. 

(c) Acceleration vector: f = q = xi + yj + zk. 

IV. Basic laws of mechanics. 

(a) Law of motion: P = raf, (P = force). 

(b) Law of action and reaction: Action and reaction are equal 
and opposite. 

(c) Law of the parallelogram of forces : P + Q is the resultant 
of P and Q. 

EXERCISES I 

1. If two forces of magnitude P and Q act at an inclination to one 
another, prove that the magnitude of the resultant R is given by 

7t!2 = P2 + Q2 + 2PQ cos B. 

2. A particle is acted on by forces of magnitudes P and Q, their lines of 
action making an angle with one another. They are to be balanced by 
two equal forces, acting at right angles to one another. Find the common 
magnitude of these forces. 

3. Show that if the magnitudes of a number of coplanar vectors are 
multiplied by a common factor, the directions of the vectors being 
unchanged, the magnitude of the sum is multiplied by the same factor and 
its direction is unchanged. Show also that if all the vectors are rotated 
through a common angle in their plane, without change of magnitude, their 
sum is rotated through the same angle without change of magnitude. 

4. Forces of magnitudes 3, 4, and 5 Ib. wt. act at a point in directions 
parallel to the sides of an equilateral triangle taken in order. Find their 
resultant. 

5. Two forces acting in opposite directions on a particle have a resultant 
of 34 Ib. wt.; if they acted at right angles to one another, their resultant 
would have a magnitude of 50 Ib. wt. Find the magnitudes of the forces. 

6. An airplane dives at 400 miles per hour, losing height at the rate of 
220 ft. per sec. What is the horizontal component of its velocity in miles 
per hour? 

7. Coplanar forces of magnitudes P, 2P, 4P act on a particle. How 
should they be directed to make the resultant (i) a maximum, (ii) a min- 
imum? 

8. If any number of coplanar vectors all of the same magnitude are 
drawn from a point, arranged symmetrically so that the angles between 
adjacent vectors are all equal, prove that their sum is zero. 

9. If V = x* + y* + z 2 -f xy + x, at what points in space is the vector 
grad V parallel to the z-axis? 

10. A scalar field is given over a plane by 

i, - *' + y 



Ex. II FOUNDATIONS OF MECHANICS 37 

What are the level curves? Show that, at the point with polar coordi- 
nates (r, 0), grad V is inclined to the re-axis at an angle 20 and its magnitude 
is $ sec 2 0. 

11. A train, starting at time t = 0, has moved in time t a distance 

x = at(l - e- 6 '), 

where a and 6 are positive constants. Find its velocity and acceleration; 
what do these become after a long time has elapsed? 

12. A particle travels along a straight line with constant acceleration /. 
Prove 

.s = ut + %ft*, v = u + ft, v* = u z + 2/s, 

where s is the distance covered from the instant t = 0, u the initial velocity, 
and v the final velocity. 

13. A uniformly accelerated automobile passes two telephone poles with 
velocities 10 m.p.h. and 20 m.p.h., respectively. Calculate its velocity 
when it is halfway between the poles. 

14. What curve is described by a particle moving in accordance with the 
equation 

r = a cos ct i + b sin ct - j, 

where a, 6, c are constants arid i, j fixed unit vectors perpendicular to one 
another? Show that the acceleration is directed toward the origin. 

15. An elevator weighing one ton starts upward with constant accelera- 
tion and attains a velocity of 15 ft. sec." 1 in a distance of 10 ft. Find in 
tons weight the tension in the supporting cable during the accelerated 
motion. 

16. A car weighing 2 tons comes to rest with uniform deceleration from a 
speed of 30 m.p h. in 100 ft. Find the force exerted by the car 011 the road, 
showing its direction in a diagram. 

17. A particle of mass m moves on the axis Ox according to the equation 

x =* a sin pt, 

where a and p are constants. Express the force acting on it as a function 
of x. 

18. A tug tows a barge A, which in turn tows another barge B. They 
start to move with an acceleration /. Find the tensions in the towing 
cables, in terms of / and m, m f (the masses of the barges). 

19. Indicate the fallacy in the following argument: A locomotive pulls 
a train. But to every action there is an equal and opposite reaction. 
Therefore the train pulls the locomotive backward with a force equal to the 
pull of the locomotive, and so there can be no motion. 

20. If the fundamental law of mechanics for a particle moving on a 

straight line were 

d ( 



instead of (1.402), m and c being constants, find the distance traveled from 
rest in time t under the action of a constant force P. (This is the relativistic 
equation of motion; see Chap. XVI.) 



CHAPTER II 

METHODS OF PLANE STATICS 
2.1. INTRODUCTORY NOTE 

In order to deal systematically with the subject of mechanics, 
we have to break it up into parts. The first division is into 
statics and dynamics: statics deals with the equilibrium of systems 
at rest, and dynamics with the motion of systems. As we have 
already seen, rest and motion are terms which have meanings 
only when a frame of reference has been specified. Thus the 
question at once arises: When we say that statics deals with 
systems at rest, what physical frame of reference have we in 
mind? The mathematical theory of statics is based on the 
fundamental laws of Sec. 1.4. Thus we should be confident of a 
close agreement between theory and observation if we were to 
develop statics relative to the astronomical frame. But that 
would not be physically interesting, because there are actually 
no systems at rest in that frame; the earth's surface is the 
physically interesting frame of reference for statics. Although 
(1.401) is not satisfied with great precision relative to the earth's 
surface, it is possible by a modification of the forces (i.e., by 
inclusion of centrifugal force) to obtain extremely satisfactory 
results in statics relative to the earth's surface. Thus in the 
mathematical theory of statics the frame of reference will be such 
that the fundamental laws hold, and in the physical interpreta- 
tion of the results the frame of reference will be the earth's 
surface. 

But there is another division of the subject of mechanics, 
namely, a division into plane mechanics and mechanics in space. 
This division is artificial from a physical point of view but is 
convenient in learning the subject, because the mathematics 
of the plane theory is simpler than the mathematics of the space 
theory. This is due to the fact that certain quantities (moments 
of forces, angular velocity, and angular momentum) appear as 
scalars in the plane theory but as vectors in the space theory. 

38 



SBC. 2.2) METHODS OF PLANE STATICS 39 

Accordingly, to avoid undue mathematical complications in the 
development, we shall deal first with plane mechanics, but where 
the space theory presents no difficulty we shall develop it 
simultaneously. 

To be precise, the subject of plane mechanics deals with 
(i) The statics and dynamics of a system of particles lying in a 
fixed plane. 

(ii) The statics and dynamics of rigid bodies which can move 
only parallel to a fixed plane, the displacement or velocity of 
every particle being parallel to the fixed plane. 

As an example of (i), we may mention the problem of the 
motion of the earth relative to the sun, both being treated as 
particles, and as an example of (ii) the motion of a cylinder 
rolling down an inclined plane. 

The plane in which the system lies, or to which the motion is 
parallel, will be called the fundamental plane. 

We shall find in both statics and dynamics that the basic 
laws lead to certain general principles or methods. When 
any specific problem presents itself, we do not attack it directly 
from first principles as a rule; we can avoid waste of energy by 
applying to the problem one of the general methods. The 
present chapter is devoted to general methods in plane statics, 
with inclusion of the space theory where it presents no difficulty. 

2.2. EQUILIBRIUM OF A PARTICLE 

According to (1.401) a particle will have an acceleration unless 
the force acting on it vanishes. Thus if P is the force acting 
on a particle, the necessary and sufficient condition for equilib- 
rium is 

(2.201) P = 0. 

If several forces P, Q, R act on a particle, the necessary 
and sufficient condition for equilibrium is the vanishing of the 
resultant force, i.e., 

(2.202) P + Q + R+ ' ' =0. 

This vector condition may also be expressed in scalar form by 
means of components. Let a particle be acted on by forces 
whose components on an orthogonal triad are indicated as 
follows: 



40 PLANE MECHANICS [Sac. 2.2 

P(P 1 ,P 2 ,P,), 

Q, Oi), 
#2, Ra), 



Then the conditions for equilibrium are 

(Pi + 0i + fii+ -o, 

(2.203) ) P 2 + Q 2 + #2 + = 0, 

(P, + Q3 + Rs + ' = 0. 

All the above remarks hold whether the forces acting on the 
particle lie in a plane or not. The particular feature of the plane 
case is that only two components are to be considered instead of 
three. 

No difficulty will be found in using (2.202) to prove the follow- 
ing theorems which are often useful : 

(i) THE TRIANGLE OF FORCES. If a particle is in equilibrium 
under the action of three forces, these forces may be represented 
in magnitude and direction by the three sides of a triangle, taken 
in order (Fig. 9). 





R 

Forces on patficle "V Q Triangle of forces 



Fio. 9. The triangle of forces. 



(ii) THE POLYGON OF FORCES. If a particle is in equilibrium 
under the action of several forces, these forces may be repre- 
sented by the sides of a closed polygon, taken in order. 

(iii) L AMY'S THEOREM. If a particle is in equilibrium under 
the action of three forces P, Q, R, then 

P __ Q __ R 



(2.204) 



sin a sin ft sin 7 



where a is the angle between Q and R, ft the angle between R and 
P, and 7 the angle between P and Q. 

Exercise. A particle is in equilibrium under three forces. Two of the 
forces act at right angles to one another, one being double the other. The 



SEC. 2.3] METHODS OF PLANE STATICS 41 

third force has a magnitude 10 Ib. wt. Find the magnitudes of the other 
two. 

2.3. EQUILIBRIUM OF A SYSTEM OF PARTICLES 
Systems of particles. 

Let us now consider a system of particles. We must in all 
cases come to a clear understanding as to what is included in the 
system under consideration. Let us suppose that we are dealing 
with a book which rests on a table, the table standing on the 
floor. The book and the table are regarded (in our mathematical 
model) as composed of a very great number of particles. In 
talking about this arrangement of matter, we are not compelled 
to think of the "system" under consideration as composed 
of the book and the table. If we like, we may think of the table 
alone as a system, or the book alone as a system, or the hundredth 
page of the book as a system. 

It is important to realize that the system under consideration 
is something we pick out from the given arrangement in an 
arbitrary manner. It is necessary to understand this in order 
to appreciate the distinction between external and internal forces. 

External and internal forces. 

The book presses down on the table and the table presses 
up on the book with an equal and opposite force (law of action 
and reaction). If the system is book + table, these forces are 
both exerted by particles of the system. But if the system 
consists of the book only, this is no longer the case; the force 
exerted by the table on the book is due, not to particles of the 
system, but to particles lying outside the system. 

We make the following general definition: 

A force acting on a particle of a given system is an internal 
force when it is exerted by another particle of that system; 
otherwise it is an external force. 

In accordance with the law of action and reaction, internal 
forces occur in equal and opposite pairs, each pair representing 
the mutual interactions of a pair of particles of the system. 

Figure 10 shows three particles A, B, C, the broken line indi- 
cating the boundary of the system. The forces are classified 
as follows: 



42 



PLANE MECHANICS 



ISnc. 2.3 



External: P at A, Q at B, R at C. 
Internal: U at B, -U at C, V at (7, -V at A, 
Wat A, -Wat B. 

Figure 11 shows these particles situated precisely as in Fig. 10 
and subject to the same .forces. The only difference between the 
two diagrams lies in the position of the broken line, indicating 
the -boundary of the system under consideration. In Fig. 11 the 
system contains only the particles A and B. Now P at" A, 
- V at A, Q at B, U at B are external; and W at A, W at B are 
internal. If we reduce the system under consideration to one 
particle only, there are no longer any internal forces. 




FIG. 10. External and internal 
forces. The "system" is enclosed 
by the broken line. 



FIG. 11. The same particles and 
forces as in Fig. 10, but a different 
"system." 



The only essential difference between the treatment of this 
question of internal and external forces in a plane and its treat- 
ment in space is that in the former case we may delimit the 
system by a closed curve, whereas in the latter case we require a 
closed surface. 

Necessary conditions for equilibrium (forces). 

We shall now obtain necessary conditions for the equilibrium 
6f any system of particles. The meaning of the word " neces- 
sary" should be emphasized: these conditions must be satisfied if 
the system is in equilibrium, but the satisfaction of the conditions 
does not imply that there is equilibrium. 

Consider any particle of the system, assumed in equilibrium. 
That particle is itself in equilibrium, and hence the vector sum of 
all forces acting on it is zero. Similarly for all particles. Hence, 

The vector sum of all forces acting on all particles is zero. 
But the forces are some external, some internal. We may state 



SEC. 2.3] METHODS OF PLANE STATICS 43 

The vector sum of all external forces and all internal forces is zero. 
From the equality of action and reaction, 

The vector sum of all internal forces is zero. 
Comparing this with the preceding statement, we have 

(2.301) The vector sum of all external forces is zero. 

Here we have the first of the necessary conditions for the equilib- 
rium of a system of particles. 

Exercise. Consider a glass of water standing on a table, taking as the 
system (i) the water only, (ii) the water and the glass. What are the 
external forces in each case, and what does (2.301) tell us about them? 

The moment of a vector about a line. 

Consider a line L and a bound vector P, perpendicular to L 
but not intersecting it. Let a be the length of the common 



L 




L 




P 

A 



FIG. 12. A line L and a 

vector P perpendicular to it. FIG. 13. (a) Af 
The moments in the two cases 
have opposite signs. 

perpendicular to L and the line of action of P. The moment of 
P about L is defined to be 
(2.302) M = aP. 

The ambiguous sign is introduced in order that we may distin- 
guish between the two cases shown in Fig. 12, in which the vector 
P indicates rotations in opposite senses about L. Having decided 
to use the + sign for vectors indicating rotations in one sense, 
we use the sign for vectors indicating rotations in the opposite 
sense. The significance of the signs will be better understood 
when Sec. 9.3 has been read, but for the present the following 
description of the convention will serve. 

Let us suppose the line L draAvn perpendicular to the plane of the 
paper, intersecting it at the point A (Figs. 13a and b) ; the moment 
is +aP when P indicates a counterclockwise rotation around A 
in the plane, and aP when a clockwise rotation is indicated. 

Consider now a line L and a bound vector P which is not per- 
pendicular to L. The moment of P about L is defined to be the 



44 



PLANE MECHANICS 



[SEC. 2.3 



moment about L of the projection of P on a plane perpendicular 
to L, the latter moment having already been defined above. 

The following facts are now obvious : 

(i) The moment of P about L is unaltered if P is made to slide 
along its line of action without change of magnitude or sense. 
(It is seen at once that this only slides the projection along its 
line of action, without change of magnitude or sense.) 

(ii) The sum of the moments about any line L of two vectors 
P, P situated on the same line is zero. (Their moments are 
equal in magnitude but opposite in sign.) 

(iii) The moment about L of a vector P is unaltered if P is 
moved without change of magnitude or direction in a direction 
parallel to L. (This does not change its projection on a plane 
perpendicular to L.) 

When we deal with the moments of coplanar vectors about a 
line perpendicular to their plane, we often refer to these moments 

as moments about a point, namely, 
the point where the line cuts the 
plane. 

The theorem of Varignon. 

Consider a line L and a bound 
vector R at a point B (Fig. 14). 
Let A be the foot of the perpendic- 
ular dropped from B on L. Let Q 
be the projection of R on the plane 
through B perpendicular to L. Let 
N bo the line through B perpendic- 
ular to AB and to L, and let P be 
the projection of Q on N. It is 
clear that P is also the projection of 

R on N. We wish to prove that the moment of R about L is 

equal to the moment of P about L. 

Figure 15 shows the plane containing P, Q, and AB. Let AC 

be drawn perpendicular to the line of action of Q, and let the 

angle CAB be denoted by B. Then the moment M of R about L 

equals the moment of Q about L (by definition), so that* 
M = Q - AC = Q AB cos = AB Q cos 6 = AB - P, 

which is the moment of P about L. Hence, the moment of a vector 
* For simplicity, we have taken the case where M is positive; the other 

case is dealt with similarly. 




Q 



FIG. 14. By definition, Q 
and R have the same moment 
about L\ it is to be proved that P 
also has the same moment. 



SEC. 2.3] 



METHODS OF PLANE STATICS 



45 



at B about a line L is equal to the moment about L of the projection 
of the vector on the line through B perpendicular to the plane con- 
taining B and L. f 

Since the projection on any line of the sum of any number of 
vectors is equal to the sum of their projections on that line, the 
theorem of Varignon follows immediately: 

The sum of the moments about a line L of vectors P, Q, R, , 
with common origin B, is equal to the moment about L of the single 
vector P + Q + R -f - with origin B. 




FIG. 15. Tho plane contain- 
ing P and Q. 



O 

Fia. 16. Analytical method of 
finding the moment of a vector. 



In statics the only vectors whose moments we have occasion to 
consider are forces. But it should be noted that the above 
definitions and theorems hold for any vectors. We shall have 
occasion to use them in Chap. V in discussing angular momentum. 

The following analytical result is important; it is obvious from 
Fig. 16, which shows the projection on the plane Oxy. If a 
vector with components (X, Y, Z) acts at the point (x, y, z), then its 
moment about the perpendicular to the plane Oxy at is 

(2.303) M = xY - yX. 

The moment about the perpendicular to the plane Oxy at the 
point (a, 6) is 

(2.304) M - (x - a)Y - (y - b)X. 

These formulas take care of the sign of M automatically. 

Exercise. A force, of fixed magnitude R and variable inclination B to the 
z-axis, acts in the plane Oxy at the fixed point (a, 6). Find its moment 
about the origin as a function of 0, and obtain the values of for which this 
moment (i) is a maximum, (ii) is a minimum, (iii) vanishes. 

Necessary conditions for equilibrium (moments). 

Let us now return to the consideration of a system of particles 
in equilibrium. Let L be any line. Consider any particle. 
Since it is in equilibrium, the resultant of all the forces acting on 



46 PLANE MECHANICS [SEC. 2.3 

it is zero; hence, by Varignon's theorem the sum of the moments 
about L of all forces acting on the particle is zero. Similarly 
for all particles. Hence, 

The sum of the moments about L of all external forces and all 
internal forces is zero. 

But the sum of the moments of a pair of equal and opposite 
internal forces is zero. Hence, 

The sum of the moments about L of all internal forces is zero. 
Comparing this with the preceding statement, we have 

(2.305) The sum of the moments about L of all external forces is 

zero. 

From (2.301) and (2.305) we may now state necessary condi- 
tions for the equilibrium of any system of particles. 

// a system of particles is in equilibrium, then 

(i) The vector sum of all EXTERNAL forces is zero. 

(ii) The sum of the moments of all EXTERNAL forces about 
any line is zero. 

This result is the key to the solution of statical problems, and 
it should be remembered. 

The particular form of the above statement applicable to 
plane statics is as follows: 

// a system of particles is in equilibrium, then 

(2.306) F = 0, N - 0, 

where F is the vector sum of the projections of all EXTERNAL 
forces on the fundamental plane, and N the sum of the moments 
of all EXTERNAL forces about any line perpendicular to the 
fundamental plane. 

It is convenient to have an explicit form of (2.306). Let axes 
be chosen so that the fundamental plane is z = 0. We shall take 
moments about the z-axis. Suppose now that the system is 
acted on by external forces with components (Xi, YI, Zi), 
(X*, 7 2 , Z 2 ), (X n , Y n , Z n ) at points (xi, y\, Zi), (x*, y z , z 2 ), 
(x, 2/n, Zn). By projection on the fundamental plane, it 
follows that F and N are unaltered if we replace this system.by 
forces in the fundamental plane with components (Xi, FI), 
(X 2 , Y 2 ), - - (X n , Y n ) at points (xi, ?/i), (#2, 2/2), (x n , 2/). 
Hence, by (2.303) we see that if (X, Y) are the components of F, 
then 

(2.307) X - 2 X t , Y = 2 K t , N = 5 (x % Yi 



SBC. 2.3) METHODS OF PLANE STATICS 47 

and so necessary conditions for equilibrium are 



(2.308) Xi = 0, V Y t = 0, V (.r t r< - yXj - 0. 
<-i 1=1 t~i 

It might be thought that, by taking moments about other 
lines perpendicular to the fundamental plane, new conditions 
might be obtained; but this is not so. For, by (2.304), if moments 
are taken about a perpendicular to the plane at (a, 6), then the 
total moment is 

- (y. - 6) AM = V (x.Y< 



and this vanishes automatically if (2.308) are satisfied. Thus 
conditions of the type (2.306) or (2.308) are actually three in 
number, and no more. 

The following important results are easy to establish from the 
principles laid down above: 

(i) If a system is in equilibrium under the action of only two 
external forces, then these forces have a common line of action, 
equal magnitudes, and opposite senses. 

(ii) If a system is in equilibrium under the action of only throe 
external forces, these forces lie in a piano and thoir lines of action 
are either concurrent or parallel. 

If a force is measured in dynamical units, its moment has dimensions 
[ML 2 ? 1 " 2 ] (see Appendix). In the c.g s. system, the unit moment is 1 
dyne em. or 1 gm. cm. 2 sec." 2 ; in the f.p.s. system, it is 1 ft. poundal or 1 Ih. 
ft. 2 see." 2 In statics we frequently use a unit of forco which is not a dynami- 
cal unit, such us the Ib. wt. or the ton wt. (contracted to read Ib. and ton). 
The corresponding moments are measured in ft. Ib. or ft. ton. 

Equipollent systems of forces. 

Two systems of forces are said to be equipollent* when the 
following conditions are satisfied: 

* The word equivalent is often used. But there is a danger of confusion 
in using a common word in a technical sense. We might think that the 
effects of two such force systems were the same; this is not always the case. 
If we pull a string with equal and opposite forces at its ends, we produce a 
very different effect from that caused by pushing it with these forces reversed 
in direction; yet the two force systems are equipollent. 



48 PLANE MECHANICS [Sic. 2.3 

(i) The vector sum of all the forces of one system is equal to the 
vector sum of all the forces of the other system. 

(ii) The sum of the moments of all the forces of one system 
about an arbitrary line is equal to the sum of the moments of 
all the forces of the other system about that line. 

For the discussion of plane mechanics, we require only a 
restricted type of equipollence, which we shall call plane equi- 
pollence. Two systems of forces are said to be plane-equipollent 
for the fundamental plane if 

(i) The vector sum of the projections on the fundamental 
plane of all the forces of one system is equal to the vector sum 
of the projections on that plane of all the forces of the other 
system. 

(ii) The sum of the moments of all the forces of one system 
about an arbitrary line perpendicular to the fundamental plane 
is equal to the sum of the moments of all the forces of the other 
system about the same line. 

It is evident that if F, F' are the vector sums of the projections 
of the forces of the two systems, and N, N' the moments about 
some line, then the conditions for plane equipollence are 

(2.309) F = F', N = N'. 

If F = 0, N = 0, we say that the system is plane-equipollent to 
zero. Thus, if a system of particles is in equilibrium, the forces 
acting on it are plane-equipollent to zero. 

The idea of equipollence is extremely useful in statics. The 
solutions of problems in plane statics turn on the conditions 
(2.306), and difficulties may arise in the calculation of F and N. 
These difficulties are reduced by splitting up the calculation into 
parts, each of which is simple. This reduction depends on the 
following fact (obvious from the definition of equipollence) : For the 
calculation of the vector sum of the projections of the forces of a sys- 
tem on the fundamental plane and the sum of their moments about a 
line perpendicular to that plane, any system of forces may be replaced 
by a system plane-equipollent to it. 

In what follows below, we shall speak only of forces in the 
fundamental plane. This makes for simplicity of expression 
without any real loss of generality, because in problems of plane 
mechanics we are actually interested in projections on the funda- 
mental plane and moments about lines perpendicular to it; 



SBC. 2.3] METHODS OF PLANE STATICS 49 

for the calculation of these, we may replace a given force system 
by a plane-equipollent system in the fundamental plane. 

Exercise. Find a system of two forces equipollent to a system of three 
forces represented by the sides of an equilateral triangle taken in order. 

Couples. 

A couplo is defined as a pair of forces acting on parallel lines, 
equal in magnitude and opposite in sense. In fact, a couple 
consists of a pair of forces P, P. A line drawn perpendicular 
to the two lines of action and terminated by them is called the 
arm of the couple. 

A A 





17cr. 176. 

FIG. 17. (a) Couple with positive moment, (b) Couple with negative moment. 

Thus the vector sum of the forces constituting a couple is zero. 
Consider now the sum of the moments of the forces forming a 
couple about any line L, perpendicular to the plane of the couple 
and cutting it at A (Figs. I7a and 17fr). If AB, AC arc the 
perpendiculars dropped from A on the lines of action, the sum 
of the moments is 

M = P - AC - P AB = P BC = Pa (Fig. 17a), 

M = P - AB - P AC = -P BC = -Pa (Fig. 176), 

where a is the arm of the couple. The rule for sign is easily seen 
to be as follows: 

(2.310) M = Pa, 

where the + or sign is to be taken according as the forces 
indicate a positive (counterclockwise) rotation or a negative 
(clockwise) rotation in the plane of the couple about any point 
taken between their lines of action. 

We observe that the sum of the moments of the forces forming 
a couple about a line perpendicular to its plane is the same for all 



50 PLANE MECHANICS [SBC. 2.3 

such lines. Hence, we may speak in an absolute sense of the 
moment of a couple. Since the vector sum of forces in a couple is 
zero, it follows that two couples in the fundamental plane are 
plane-equipollent if they have the same moment. 

It is evident that two couples of moments M, M r in the funda- 
mental plane are together plane-equipollent to a single couple of 
moment M + M r in that plane. 

Reduction of a general plane force system. 

Consider a system of forces in the fundamental plane. Let 
F be their vector sum and N the sum of their moments about 
some point in the plane. * Consider on the other hand a single 
force F at and a couple of moment N in the fundamental plane. 
Obviously, this simple system is plane-equipollent to the given 

system. Hence, we may state 
the following general result: 

A system of forces in the funda- 
mental plane is plane-equipollent 
to a single force applied at an arbi- 
trary point in the plane, together 
with a couple. 

FIG. 18. Reduction of a force and a It is customary also to express 

couple to a single force. ., . , . ,, . f 

this by saying that a system of 

forces in the fundamental plane may be reduced to a force and a 
couple. 

Expressed analytically, a system of forces (Xi, Fi), (X z , F 2 ), 
- (X n , F n ) at points (xi, yi), (x 2 , 2/2), (x, y) may bo 
reduced to a single force at the origin with components X, F, 
together with a couple N 9 where 

(2.311) X = X i9 Y = F t , N 

We shall now show that a still greater reduction is possible. 

Let be an arbitrary point. The force system, as we already 
know, may be reduced to a force F at 0, together with a couple 

of moment N. In Fig. 18, the force F at is shown as OB. We 
now consider the following two cases: 

* That is, the moments about a line perpendicular to the fundamental 
plane, cutting it at O (see p. 44). 




SBC. 2.3] METHODS OF PLANE STATICS 51 

CASE (i): F 7* 0. Let us suppose N positive for simplicity, 
the case where N is negative being similarly dealt with. We 
draw OA perpendicular to OB as in Fig. 18 and measure off OA 
equal to N/F. Then the couple is equipollent to the pair of forces 

OC (or -F at 0) ; AD (or F at A). 
Thus the given system is reduced to the three forces 
F at 0, -F at 0, F at A. 

These are equipollent to a single force F at A. 

CASE (ii) : F = 0. Here the system is reduced to a couple. 

Hence we may state the following general result : 

Any force system in the fundamental plane may be reduced either 
to a single force or to a couple. 

It may be noted that reduction to a single force will occur much 
more frequently than reduction to a couple, because reduction 
to a couple occurs only when a special condition is satisfied, 
viz., F = 0, or, in other words, when the vector sum of the 
forces in the system is zero. 

Let us now carry out this reduction to a single force or to a 
couple analytically, the given plane force system being specified 
as consisting of forces with components (Xi, FI), (X^ Y%), 
(X n , Y n ) at points with coordinates (xi, y\\ (#2, 2/2), (#n, 2/n). 
Let us suppose that this system may be reduced to a force 
with components (X, F) at (x t y). The conditions of plane 
equipollence are 

(2.312) X = 2* X^ F = 2) F t , 

n 

xY z/X= V ( x *^ " 

The first two equations determine the components of the single 
force. The last equation gives one relation between the coordi- 
nates of the point of application; this relation, being linear, 
defines a straight line. It is seen at once that this line points 
in the direction of the force with components (X, F); it is, in 
fact, the line of action of that force. The last equation of 
(2.312) leaves the point of application indeterminate to a certain 
extent it may take any position on a certain line. One particu- 



52 PLANE MECHANICS [Sue. 2.3 

lar point on this line is given by the symmetrical formulas 



(2.313) x = -=- - , y 



If it should happen that 

(2.314) j\ X> = 0, V 7, = 0, 2 feF< - yA) ^ 0, 
t=i i=I ~ 

it is evident that we cannot find Z, F ; a;, y to satisfy (2.312). 
Then the system cannot be reduced to a single force. To find 
the couple to which it can be reduced, we compare it with the 
couple consisting of forces (0, 7), (0, F) at the points (0, 0), 
(0, x\ respectively. The conditions of equipollence are 



+ = 0, -7 + F = 0, xY = 



= V 

i = l 



These equations are satisfied provided that the moment of the 
couple is 

(2.315) M = (x,y. - 2/ t X t ). 

To sum up: In general a plane system of forces is plane-equi- 
pollent to a single force whose components and line of action are 
given by (2.312). // 

(2.316) iX = 0, 



the system is plane-equipollent to a couple with moment given by 
(2.315). // (2.316) are satisfied and also 

(2.317) 2} (x t y f - yt Z t ) - 0, 

then the given system is plane-equipollent to zero. 

Exercise. Forces of magnitudes 2 and 3 act parallel to the x-axis at points 
(1, 3) and (2, 4), respectively. Reduce them (i) to a force at the origin and 
a couple, (ii) to a single force. 




SEC. 2.4] METHODS OF PLANE STATICS 53 

2.4. WORK AND POTENTIAL ENERGY 
Definition of work. 

Consider a particle A on which a force P acts. Let the particle 
be given an infinitesimal displacement of magnitude 5s, repre- 
sented by AB (Fig. 19). The work done by P in this displace- 
ment is defined to be the product of 5s and the component of 
P in the direction of the displacement; in fact, the work 5W is 

(2.401) 8W = P cos 6 - 5s, 

where 6 is the angle between P and the displacement.* It will 
be positive or negative according as 
Q is acute or obtuse. A 

Since the sum of components in any 
direction is equal to the component of 

,, . ,, , ,. ,. ., Fia. 19. The force P does 

the sum in that direction, it follows work whc n the particle on which 
that the total work done by any num- it acts receives the displacement 
ber of forces P, Q, R, , acting on 

a particle, is equal to the work done by their resultant P + Q 
+ R+ .-.. 

It is evident that 8W is also equal to the product of P and the 
component of the displacement in the direction of P. Hence 
the work done by P in a succession of small displacements is 
equal to the work done by P in the resultant displacement. 

Let X, Y, Z be the components of P in the directions of the 
axes of coordinates, and let the coordinates of A, B be (x, y, z), 
(x + dx, y + dy, z + 5z), respectively. Then, since the direction 
cosines of P are X/P, Y/P, Z/P and those of the displacement 

AB are 5z/5s, 5?//5s, 5s/ 5s, we have 

X dx , F5?/ . Z dz 

(2.402) cos Q = -D -r- + -Q -r + D T-> 

t os i os i 5s 

and so the work done by P in the displacement AB is 

(2.403) 8W = X dx + Y Sy + Z 5z. 

* The use of the symbol 5 instead of the more usual d (for differential) is 
traditional, and not of much importance as far as statics is concerned. But 
in dynamics it is necessary to distinguish between a purely hypothetical 
(or virtual) displacement dx and the displacement dx actually occurring in 
time dt (i e., dx x dt). 



54 PLANE MECHANICS [Ssc. 2.4 

We note that no work is done when the displacement is 
perpendicular to the force. 

If force is measured in dynamical units, work has dimensions [AfL 2 jT~ 2 ] 
(see Appendix). In the c.g.s. system, the unit of work is the erg, which is 1 
dyne cm. or 1 gm. cm. 2 sec." 2 ; in the f p.s. system, it is 1 ft. poundal or 1 Ib. 
ft. 2 sec*" 2 In statics, the ft. Ib. and ft. ton are used. 

Rate of working is called power. In dynamical units, power has dimen- 
sions [ML 2 r~ 3 ]. A unit commonly employed is the horsepower (550 ft. Ib. 
wt. sec."" 1 = 1 hp.). 

Forces which do no work. 

Consider a particle in contact with the surface of a fixed rigid 
body, and suppose that a force acts on the particle tending to 
drive it into the body. If no other force acted, the particle 
would have to penetrate the body, in accordance with (1.401), 
Since, however, we regard such penetration as impossible, 
we must assume the existence of another force, the reaction of 
the surface, which prevents the penetration from taking place. 
The particle in question may be an isolated particle, or it may 
be one of the particles of a rigid body. 

In a mechanical problem the reactions between particles and 
surfaces, or between pairs of surfaces, are not in general to be 
regarded as known forces. They are called into play solely to 
prevent violation of the condition of non-penetration. 

The words "smooth" and "rough" are familiar in ordinary 
life; we speak of polished steel, glass, ice, etc., as smooth, and 
sandpaper, cloth, etc., as rough. We make such a classification 
primarily on the basis of our sense of touch. A more scientific 
classification is obtained by examining the directions of the reac- 
tions between bodies. It is found that with smooth bodies the 
reactions always lie very close to the common normal of the 
surfaces in contact. As an idealization of such bodies, we admit 
into our mathematical model the concept of a smooth surface 
with the following property: 

The reaction at a smooth surface is normal to the surface, and 
is of such a magnitude as just to prevent penetration or overlapping 
in space from taking place. 

The reaction at a rough surface has more complicated proper- 
ties which will be discussed in Sec. 3.2. 

Since the reaction at a smooth surface is normal to the surface, 
the following result is evident from (2.401) : 




SBC. 2.4] METHODS OF PLANE STATICS 55 

No work is done by the reaction at a smooth fixed surface in an 
infinitesimal displacement which preserves the contact. 

In any contact there are actually two equal and opposite 
forces involved, one acting on each body. In what has been 
said above, one body was supposed fixed, so that the force acting 
on it did no work. Suppose now that both bodies, having 
smooth contact, are displaced infinitesimally. The displace- 
ments of the particles of the two bodies in contact with one 
another may now have components along the common normal, 
and so work is done by each of the two forces of reaction. But 
it is* not difficult to see that the 
sum of these two works is zero. 

Let us now consider a rolling 
contact, confining our attention 
here, for simplicity, to the rolling 
of a rigid circle on a fixed line in 
the fundamental plane (Fig. 20). __....._. 

We say that the circle C rolls A' B 

on the line L if it passes con- F ' ' 20 - A circle roUmg on 
tinuously through a sequence of positions such that (i) L is 
always tangential to C; (ii) if any two points A, B of C make 
contact with points A', B' of I/, then arc AB = A'B'. (Consider 
a motor tire and the pattern it leaves on the road.) 

If C advances a distance As while it turns through an angle 
A0, it is clear that 

(2.404) Az = a A0, 

where a is the radius of C. During this advance the point of C 
initially in contact with L receives the following displacements: 

Horizontal: Ax a sin A0, 
Vertical: a a cos A0. 

If the displacement As is infinitesimal, then these displace- 
ments are infinitesimals of the orders (Ax) 3 , (Ax) 2 , respectively. 
This is easily seen on using the well-known series for sine and 
cosine. 

Suppose now that there is a reaction R exerted by L on C. 
We shall not assume that R is perpendicular to L. The work 
W done by R in a succession of infinitesimal rolling operations 



56 PLANE MECHANICS [SBC. 2.4 

will be a function of x y the final displacement. It may be written 



But since the displacement of the point of contact is an infinitesi- 
mal of higher order than the increment in x, we have dW/dx = 
and hence W = 0. Thus, no work is done by the reaction at a 
rolling contact. It is true that our proof deals only with the case 
of a circle rolling on a line; the general case of a moving curve 
rolling on a fixed curve may be discussed by a slightly more 
complicated argument leading to the same result. The condition 
of rolling is the equality of arcs on the two curves between points 
of contact; the essential point in the proof is the fact that the 
displacement of the point of the moving curve instantaneously 
in contact is an infinitesimal of higher order than the infinitesimal 
angle through which the moving curve turns. 

When rolling takes place between two moving surfaces, the 
sum of the works done by the equal and opposite reactions is zero. 

This follows from the fact that 
the displacements of the two 
particles in contact (one belong- 
ing to each body) are equal, to 
the first order of small quantities. 
The fact that no work is done 
at a rolling contact is of enormous 
importance in modern transport, 

-Internal reactions in a which moveg Qn w hcels in COn- 
rigid body. 

trast to the dragged vehicles of 

primitive civilizations. It explains why boats are launched or 
hauled out of the water with much greater ease when placed on 
rollers, and why machinery operates more easily on ball or roller 
bearings than on plain bearings. 

Let us now consider the work done by a pair of equal and 
opposite reactions, exerted on one another by two particles of 
a rigid body, when the body receives an infinitesimal displace- 
ment. Let A, B be the positions of the two particles before 
displacement, and A', B f their positions after displacement 
(Fig. 21). The lines AB, A!B' make an infinitesimal angle with 
one another, and 

(2.405) AB - A'B', 




SEC. 2.4] METHODS OF PLANE STATICS 57 

since the body is rigid. If A Q , B Q are the projections of A', B' 
respectively on the line AB, we have obviously 

(2.406) AA + A B = AB + BB Q . 

Since the inclination of A'B' to AB is infinitesimal, AoB = A'B' 
to the first order of infinitesimals; hence, A Q B Q = AB by (2.405) 
and so (2.406) gives 

(2.407) AA = BB Q . 

If the forces on A, B arc respectively P, P, the amounts of 
work done by them are 

P - AA , -P BB , 

and the sum of these is zero. Hence, no work is done by a pair of 
equal and opposite reactions, exerted on one another by two particles 
of a rigid body. 

To sum up, we have seen that no work is done by 

(i) the reaction on a movable body in smooth contact with a 
fixed body; 

(ii) the pair of reactions at a smooth contact; 

(iii) the reaction on a body rolling on a fixed body; 

(iv) the pair of reactions at a rolling contact; 

(v) the pair of reactions between two particles of a rigid body. 

In the cases considered above, the particles forming the systems 
are not wholly free. The systems are, in fact, subject to con- 
straints, and the reactions are brought into play to prevent the 
violation of these constraints. Since these reactions of constraint 
do no work in permissible displacements, we speak of these 
constraints as workless. 

The principle of virtual work. 

Let us start by considering the simple system shown in Fig. 22. 
AB is a rigid bar; a small hole is drilled in it at C, and a smooth 
pin passes through the hole, attaching the bar to some fixed 
support (not shown). Thus the bar can turn about C in the 
plane of the paper. Forces P and Q are applied at A and B, 
respectively, in directions perpendicular to A B. 

There are two ways of regarding this system: 

(i) It is just a rigid body, which can turn round C and which 
is subjected to the forces P and Q. 



58 PLANE MECHANICS [SEC. 2.4 

(ii) It is a collection of a vast number of particles, subjected 
not only to the forces P and Q, but also to a vast number of 
reactions between the particles and a reaction at C, all adjusted 
so that the distances between the particles remain constant and 
the particles near C do not move away from the pin. 

For present purposes we regard (ii) as the more useful view. 
In developing the principle of virtual work, we have to deal 
with displacements, forces, and work. 

The only displacement consistent with the constraints is a 
rotation around C. But if we take the point of view (ii) above, 

we may think of other displace- 
B ments in which the constraints 
are violated for example, only 
one particle of the bar might be 
moved from its position. Such 
a displacement is merely a 
mathematical device. In either 

Fm.22.-A rigid bar pinned at C. CMe (whether the constraints 

are satisfied or not) the displace- 
ment is called "virtual/ 7 this word implying that the displace- 
ment is a hypothetical one and not a displacement actually 
experienced. We note then that virtual displacements are of 
two types: 

(a) virtual displacements satisfying the constraints, 

(6) virtual displacements violating the constraints. 

As for forces, we have P and Q and the reactions of constraint. 
We need a word to distinguish P and Q from the latter; we 
cannot use the word "external" because the reaction exerted by 
the pin at C is external. So we shall call P and Q applied forces, 
with this general definition for any system with workless con- 
straints : Forces other than reactions of constraint are called applied 
forces. Thus the forces acting on the system are of two types: 

(a) applied forces, 

(&) reactions of constraint. 

As for the work done in a virtual displacement (called virtual 
work), we are to observe that no work is done by the reactions 
of constraint provided that the constraints are of the workless 
type and are satisfied by the virtual displacement. 

We shall now proceed to state and prove the principle of 
virtual work. The argument will be quite general, covering 



SBC. 2.4] METHODS OF PLANE STATICS 59 

the case of any system in which the constraints are of the workless 
type. 

PRINCIPLE OF VIRTUAL WORK. A system with workless con- 
straints is in equilibrium under applied forces if, and only if, zero 
virtual work is done by the applied forces in an arbitrary infinitesi- 
mal displacement satisfying the constraints. 

It will be noticed that this statement contains both a suffi- 
cient (if) condition for equilibrium and a necessary (only if) 
condition. There are two .theorems here, requiring separate 
proofs. 

Let us first prove the necessity of the condition; that is, we 
are given that the system is in equilibrium, and we have to prove 
that zero virtual work is done by the applied forces in any 
infinitesimal displacement satisfying the constraints. Consider 
any particle of the system; it is in equilibrium, and so the result- 
ant of all forces acting on it is zero. Thus, zero virtual work is 
done by the forces acting on that particle in any displacement of 
it. This holds for all particles; and so, in any displacement of 
the system, zero virtual work is done by all forces acting. But 
the reactions of constraint do no work in any displacement which 
satisfies the constraints. Hence the applied forces do no work in 
such a displacement, and so the necessity of the condition is 
proved. 

Let us now prove its sufficiency; that is, we are given that 
zero virtual work is done by the applied forces in any infinitesi- 
mal displacement satisfying the constraints, and we have to 
prove that the system is in equilibrium. Suppose it is not in 
equilibrium; then it starts to move. It is clear from (1.401) 
that each particle starts to move in the direction of the resultant 
force acting on it. Referring to (2.401), we see that a positive 
amount of work is done in the initial displacement, since = 
and cos = 1. This is true for every particle; and so, in the 
initial displacement of the system, positive (not zero) virtual 
work is done by all the forces. But such an initial displacement 
must of course satisfy the constraints, and so the reactions of 
constraint do no work. Thus, on the basis of our assumption 
that the system is not in equilibrium, we see that it must 
undergo a displacement which satisfies the constraints and in 
which the applied forces do positive work. But this contradicts 
the given information, according to which zero work is done. 



60 PLANE MECHANICS [SEC. 2.4 

Since our assumption leads to a contradiction, it must be false, 
and so the system does remain in equilibrium. The sufficiency 
of the condition is proved. 

Returning to the system shown in Fig. 22, let us give the bar 
a rotation about C in a counterclockwise sense through an 
infinitesimal angle 60. The work done by P is Pa 50 and the 
work done by Q is Qb 50, and so the total work is 

6W = (Pa - Qb) 80. 
If the system is in equilibrium, then 8W = and hence 

P ^b 
Q a 

On the other hand, if P/Q = 6 /a, then 8W = for this displace- 
ment. But this is the most general displacement satisfying the 
constraints; hence the bar must be in equilibrium if the condition 
P/Q = b/a is satisfied. 

Although the chief merit of the principle of virtual work lies 
in the fact that it does not involve the reactions of constraint, 
nevertheless it can be used to find these reactions should they be 
required. Suppose, for example, we wish to know the reaction 
at C in the system considered. If this reaction is R, it Is obvious 
that the equilibrium will not be disturbed if we remove the pin 
at C and apply at C a force R. Since there is now no constraint 
at C, other virtual displacements are permissible. We may 
slide the bar along its length. In this displacement, P and Q do 
no work; hence R does no work, and consequently R acts at 
right angles to A B. If we now push the bar through a small 
distance 8x, perpendicular to A B in the upward direction, the 
work done by P and Q is (P + Q) dx. Hence the work done 
by R is (P + Q) 8x; therefore, R has a magnitude P + Q and 
acts in the direction opposite to the common direction of P 
and Q. 

We have developed the principle of virtual work for the sim- 
plest and most interesting systems, namely, those for which the 
constraints are workless. But it is easily extended to cover more 
general cases by means of the device employed above, namely, 
the replacement of a constraint by an unknown "applied" 
force. 

Exercise. A lever, in the form of the letter L, is pivoted at the angle. 
It is in equilibrium under forces applied at the ends of the arms, and per- 



SEC. 2.4] 



METHODS OF PLANE STATICS 



61 



pendicular to them. Use the method of virtual work to find the force at 
the end of one arm and the reaction at the pivot, the other force and the 
lengths of the arms being given. 

Infinitesimal displacements of a rigid body parallel to a fixed 
plane. 

Let us consider a rigid body which is permitted to move only 
parallel to a fixed fundamental plane. The section of the body 
by this plane is itself a two-dimensional rigid body; we call it 
the representative lamina. A description of the motion of this 
lamina specifies the motion of the body, and vice versa. We 
may therefore discuss infinitesimal displacements of the lamina 




B 



A 

FIG. 23.- Genoial dis- 
placement of a lamina in 
its plane. 




FIG. 24. In an infinitesimal 
displacement the quantities a, 
b, and B receive infinitesimal 
increments, while r and re- 
main constant. 



in the fundamental plane instead of displacements of the rigid 
body parallel to that plane; they come to the same thing. 

In Fig. 23, L is the lamina before and L' the lamina after an 
arbitrary displacement. Let A, B be any two points of the 
lamina before displacement and A', B' their positions after dis- 
placement. Obviously the displacement from L to L' may be 
achieved in two steps: 

(i) a translation, in which each point of L receives a displace- 

> 
ment equal and parallel to A A'] 

(ii) a rotation about A' through an angle equal to the angle 
between AB and A'B'. 

The point A, used in describing the displacement, is called a 
base point. 



62 PLANE MECHANICS [SEC. 2.4 

Let us now consider an infinitesimal displacement. For fixed 
axes Oxy in the fundamental plane (Fig. 24), let (a, 6) be the 
coordinates of A, and 6 the inclination of A B to Ox. Then 
the increments 5a, 56 describe the translational displacement, 
and the increment 86 describes the rotation. Let P be any 
particle of the lamina; let us put r = AP, <t> = BAP. Since 
the lamina is rigid, r and <j> remain constant as the lamina moves. 

Now if (x, y) are the coordinates of P, we have 

(2.408) x = a + r cos (0 + ), y = b + r sin (0 + 4). 
Hence the displacement of P in the infinitesimal displacement 
5a, 56, 50 of the lamina is 

dx = 5a - r sin (0 + <) 50, 5t/ = 56 + r cos (0 + <) 50. 
Substituting for r sin (0 + <), r cos (0 + <) from (2.408), we 
obtain 

(2.409) 8x = 5a - (y - 6) 50, 5t/ = 56 + (x - a) 50. 

This gives the infinitesimal displacement of any point in the 
lamina (or in the rigid body of which the lamina is a section) in 
terms of the translation (5a, 56) of the base point (a, 6), and the 
rotation 50. By giving arbitrary infinitesimal values to 5a, 
56, 50, we get the most general infinitesimal displacement of a 
lamina in a plane or of a rigid body parallel to a plane. 

Exercise. Find the displacement of the particle at the highest point of a 
rolling wheel, when the wheel advances a small distance 3s. 

Sufficient conditions for the equilibrium of a rigid body movable 
parallel to a fixed plane. 

We now consider the equilibrium of a rigid body which can 
move only parallel to a fixed fundamental plane. There are 
no constraints limiting the motion of the body parallel to the 
plane, but there are constraints preventing other motions. 
These are supposed to be of the workless type. In addition to 
the reactions of these constraints, there act applied forces, not 
necessarily parallel to the fundamental plane. Let (x\, 2/1), 
(^2, 2/2), * * * (x n , 2/n) be the projections on the fundamental 
plane (z = 0) of the points at which these forces are applied, 
and let (Xi, FI), (Z 2 , 7 2 ), (X n , Y n ) be the components of 
these forces in the directions of the axes Oxy. 

In an infinitesimal virtual displacement 5a, 56, 50, as described 
above, the work done is 



SBC. 2.4] METHODS OF PLANE STATICS 63 

(2.410) SW - V X t [Sa - fa - b) 38] 



= X 8a + Y db + N 50, 

where X, Y are the components of the vector sum of the applied 
forces and N is their moment about the point (a, 6). 

Thus, if 

(2.411) X = 0, F = 0, N = 0, 

we have BW = for the most general infinitesimal displacement 
consistent with the constraints. Thus, by the principle of 
virtual work, the body is in equilibrium if (2.411) are satisfied. 
These are therefore sufficient conditions for the equilibrium of 
the body. Let us restate this important result, as follows: 

// a rigid body is constrained to move parallel to a fixed funda- 
mental plane, then the body is in equilibrium under the action 
of any system of external forces plane-equipollent to zero; i.e., there 
is equilibrium provided that the vector sum of the projections of 
these forces on the fundamental plane vanishes, and the moment of 
these forces about some one line perpendicular to the fundamental 
plane vanishes also. 

We note that (2.411) are the same as (2.306) or (2.308), 
written in a slightly different form. In Sec. 2.3 these conditions 
were shown to be necessary for the equilibrium of any system; 
now we find them to be sufficient for the equilibrium of a rigid 
body movable parallel to the plane = 0. 

Let S and S' be two plane-equipollent force-systems acting 
on a rigid body; then X, F, N have the same values for S and S'. 
It follows from (2.410) that, if dW is the virtual work done by 
S and BW that done by S' in the same displacement of the body 
parallel to the fundamental plane, then 

SW = 8W. 

It is evident that two plane-equipollent force systems are 
equivalent in all statical problems concerning the equilibrium 



64 PLANE MECHANICS [SEC. 2.4 

of a rigid body movable parallel to the fundamental plane, in 
the sense that one system may be replaced by a plane-equipollent 
system without disturbing equilibrium. In particular, two forces 
are equivalent if they are equal in magnitude, with the same 
sense and with a common line of action. Thus, we may slide 
a force along its line of action without changing its effect. This 
is known as the principle of transmissibility of force for a rigid 
body, and it enables us to regard a force acting on a rigid body 
as a sliding vector. 

In the same way, two couples in the same plane are equivalent 
if they have the same moment N. Since X = Y = for a 
couple, it follows from (2.410) that the work done by a couple of 
moment N in an infinitesimal rotation 3d is N 0. 

There is no unique way in which the principles of mechanics 
must be developed. Two rival methods exist, the method of 
forces as used in Sec. 2.3 and the method of virtual work as used 
here. In developing the theory up to this stage, we have used 
one method to establish some points and the other method to 
establish other points. The reader may prefer to work out some 
other logical development of the subject, and indeed is most 
likely to appreciate the critical points in the chain of reasoning 
by so doing. 

Exercise. A card lies on a table. Along the edges, there are applied 
forces represented in magnitude and direction by the edges, taken in order. 
The card is given any small displacement. Show that the work done is 
represented by twice the area of the card, multiplied by the angle of rotation. 

Potential energy. 

Consider any system of particles. We shall refer to a set of 
positions of all the particles as a configuration of the system. 
Let AO be some configuration selected as a standard configuration, 
and let A be any other configuration. Let us take the system 
from A to Ao, and denote by W the work done by all forces 
acting on the system during this process. 

If the system consists of a single particle in the plane Oxy, 
we might take the origin as standard configuration A . Then 
if X, Y are the components of force acting on it, the work done 
in bringing it from the position A is 



(2.412) W 

the integral being taken along the curve by which the particle is 
brought to the origin. For example, if the force depends on the 



SEC. 2.4] METHODS OF PLANE STATICS 65 

position of the particle according to the equations X = 2x, 
Y = 6y and if the coordinates of A are a, 6, then 

(2.413a) W = JT* (2x dx + Qy dy) - [s + 3*/ 2 ] ;' 

=' _ a s - 352. 

This value is independent of the particular path along which 
the particle is brought to 0. 

To take another example, if X = 6y, Y = 2x, we have 

(2.4136) W = f' (Gy da: + 2x dy). 

Jo>,o 

The value of this integral is not independent of the path of 
integration, as is easily seen by taking the two paths along the 
sides of the rectangle x 0, y = 0, x = a, y = b. Thus it is 
only in some cases that W is independent of the path. 

Passing from the case of a single particle to a general system, 
we make the following definition: 

When the forces acting on a system are such that the work 
done by them, in the passage of the system from a configura- 
tion A to the standard configiiration AQ, is independent of 
the way in which this passage is carried out, then the system 
is said to be conservative. The work done by the forces in the 
passage from A to A o is called the potential energy* of the system 
at the configuration A. 

Since the standard configuration may be chosen arbitrarily, 
the potential energy of a conservative system is indeterminate 
to within an additive constant, the value of which depends on 
the standard configuration chosen. 

Potential energy will be denoted by V. Thus, in the case 
of (2.413a), the system (a single particle) is conservative and 
the potential energy V at the position x, y is 

(2.414) V = -z 2 - Si/ 2 . 

Generally speaking, most of the systems considered in me- 
chanics are conservative. The outstanding exceptions are 
systems in which frictional resistances are involved. Cases 
like that of (2.4136) occur rarely. 

Let there be a conservative system, A Q being the standard 
configuration. We shall use the following notation: 

* Since potential energy is defined as work done," it has the same dimen- 
sions [ML*T~*] and is measured in the same units as work (cf . p. 54), 



66 PLANE MECHANICS (SEC. 2.4 

V(A) = potential energy at configuration A, 
W(A, B) = work done by forces in passage from A to B. 

By the definition of 7, we have 

(2.415) V(A) = W(A, A*). 

Now give the system an infinitesimal displacement from A to B] 
let 8V be the increment in potential energy and 8W the work done. 
We have 

8V = V(B) - V(A) 



But since work done is independent of path and additive, we 
have 

W(B, A*) = W(B, A) + W(A, Ao), 
and 

W(B,A) = -W(A,B) = -ST7. 
Hence we have 

(2.416) dV = -dW. 

In words, Jta increment in potential energy equals the work done, 
with its sign changed. 

Consider a particle which can move in a plane subject to 
the action of a force which depends only on the position of the 
particle. If X, Y are the components of force and x, y the 
coordinates of the particle, then X, Y are functions of x, y. 
This is called a field of force. It is said to be a conservative field 
if the particle under its influence is a conservative system. In 
that case, denoting the potential energy by 7, we have by 
(2.403) and (2.416) the equation 

(2.417) X 5x + Y Sy - -57, 

where 6x, dy are the components of an arbitrary infinitesimal 
displacement given to the particle. It follows that 

(2.418) X = - g, Y = - f - 

The preceding result may be extended without any difficulty 
to a conservative field in space; we have then 

9V dV dV 



SBC. 2.4] 



METHODS OF PLANE STATICS 



67 



In the language of Sec. 1.3: In a conservative field, the force is the 
gradient of potential energy, with sign reversed. 

A uniform field of force is one in which X, Y, Z are constants. 
Such a field is conservative, with potential energy 



(2.420) 



V = -(Xx+Yy + Zz). 



Returning to a general conservative system, let us note a 
useful consequence of (2.416) in connection with the principle of 
virtual work. // a conservative system 
is in equilibrium, the change in potential 
energy in any infinitesimal displacement 
is zero. This is also expressed by say- 
ing that the potential energy has a 
stationary value. 

Example. As an illustrative example of the 
principle of virtual work, consider the sys- 
tem shown in Fig. 25. Two heavy particles of 
weights w, w' are connected by a light in ex- 
tensible string and hang over a fixed smooth 
circular cylinder of radius a, the axis of which 
is horizontal. We wish to find the position of 
equilibrium. 

Here we have a system of two particles. The forces acting on them are 

(i) gravity, 

(ii) forces due to the tension in the string, 

(iii) reactions exerted by the cylinder. 

If we give a virtual displacement satisfying the constraints, only gravity does 
work. The system is conservative, and the potential energy is, by (2.420), 
with suitable choice of the standard configuration, 




FIQ. 25. Two heavy par- 
ticlea balanced on a smooth 
cylinder. 



V = wa cos 6 



cos 0', 



where 0, 0' are the inclinations to the vertical of the radii drawn to the 
particles. In an infinitesimal displacement, 

dV wa sin 60 w'a sin 0' 60'. 

But + 0' is constant, since the string is inextensible. Thus 60' ** 80, 
and 

57 =s a 50(w' sin 0' w sin 0). 

Hence, when the system is in equilibrium, the following condition must be 
satisfied : 

sin _ w' 

sin 8' ~~ w 



68 PLANE MECHANICS [Sue. 2.5 

2.5. STATICALLY INDETERMINATE PROBLEMS 

Consider a rigid body which can move parallel to a fixed 
fundamental plane, in equilibrium under external forces in that 
plane. Let X and Y be the total components of the external 
forces on axes of coordinates in the fundamental plane and N 
their total moment about the origin. Then, as hi (2.411), 
we have the scalar equations of equilibrium 

(2.501) X = 0, Y = 0, N = 0. 

It is important to note that these equations are three in number. 
This means that, in any problem concerning the plane statics 
of a rigid body, we can find three unknowns and no more. If 
there are more than three unknowns, the problem is statically 



A X x P B X 2 

FIQ. 20. -A statically indeterminate problem. 

indeterminate , which means that it cannot be solved by means of 
the conditions of equilibrium alone. We shall illustrate with a 
simple example. 

A rigid bar AB (Fig. 26) is fixed at its ends; at its middle point 
there is applied a force with components P, Q along and per- 
pendicular to the bar. Find the reactions on the bar at A 
and B. 

Let the components of the reactions be Xi, Y\ at A and 
X 2 , Y z at #, and let AB = 2a. Taking components along and 
perpendicular to the bar and moments about A, by (2.501) 
we have 

( X = Xi + P + X = 0, 

(2.502) ; Y - Yi + Q + Y 2 = 0, 
( N = aQ + 2aY 2 = 0. 

Hence, 

(2.503) 7 2 = -Q, 7i = -Q, X l + X 2 = -P. 

We have only three equations for four unknowns; the problem 
is statically indeterminate, and nothing more can be found out 
about the reactions from the conditions of equilibrium alone. 



SEC. 2.6] METHODS OF PLANE STATICS 69 

If such a problem were to occur in physical reality, the four 
unknown quantities would have values which might be measured. 
Apparently our mathematical methods have failed us; they have 
not provided us with the answer to a question of physical inter- 
est. The fault actually lies in the selection of a mathematical 
model. Rigid bodies do not exist in nature, and this problem 
is one where the use of a rigid body as a mathematical model is 
not justified. We should take an elastic bar as a model instead 
(cf. Sec. 3.3). 

A simple modification in the constraints may render an 
indeterminate problem determinate. Consider the same prob- 
lem, altered by the condition that the end B, instead of being 




A X! 

FIG. 27. A statically determinate problem. 

fixed, slides on a smooth plane inclined at an angle of 45 to 
AB (Fig. 27). Again, we have (2.503), but also an additional 
equation 

(2.504) X 2 + F 2 - 0, 

arising from the condition that the reaction at B is perpendicular 
to the plane. Now the problem is statically determinate, and 
we have 

nwn 

(2.505) 

Had we taken one of the axes parallel to the plane of constraint 
at B, we should have obtained a slightly simpler treatment, 
because only three unknowns would have appeared. 

2.6. SUMMARY OF METHODS OF PLANE STATICS 
I. Conditions of equilibrium. 

(a) For single particle (necessary and sufficient) : 
(2.601) Vector sum of all forces vanishes, 



70 PLANE MECHANICS [SEC. 2.6 

or 

(2.602) Total components in two perpendicular directions 

vanish. 

(6) For any system (necessary) or for rigid body (necessary 
and sufficient) : 

(2.603) F - 0, N = 0; or X = 0, Y = 0, N = 0. 

(c) For any system with workless constraints (necessary and 
sufficient) : 

(2.604) dW = (work done by applied forces). 

II. Moment of a vector in a plane about a point in that plane. 

(2.605) M = aP (+ for counterclockwise) ; 

(2.606) M = xY - yX (about origin). 

III. Plane equipollence. 

(a) Conditions for plane equipollence: 

(2.607) F = F, N = N'. 

(6) General system of forces can be reduced to a single force 
at an assigned point, together with a couple. 

(c) General system of forces can be reduced to a single force 
or to a single couple (latter case exceptional). 

IV. Work and potential energy. 
(a) Definition of work: 

(2.608) dW = P cos - 6s = X 8x + Y By. 

(&) Reactions do no work at smooth contacts, rolling contacts, 
and inside rigid body. 

(c) For the infinitesimal displacement of a rigid body, 

(2.609) 8W = X da + Y fib + N 86. 

(d) Potential energy: 

(2.610) dV = ~dW, ~ 

(2.611) * - - ?, V=~~ 
v ' dx dy 

(Force is gradient of potential energy with sign reversed.) 



Ex. II] 



METHODS OF PLANE STATICS 



71 



EXERCISES H 

The method of virtual work may be used in any of these problems; it will 
be found particularly useful in the case of those marked with an asterisk. 

1. A particle is in equilibrium under the action of six forces. Three of 
these forces are reversed, and the particle remains in equilibrium. Prove 
that it will still remain in equilibrium if these three forces are removed 
altogether. 

2. A ladder of weight W rests at an angle to the horizontal, with 
its ends resting on a smooth floor and against a smooth vertical wall. The 
lower end is joined by a rope to the junction of the wall and the floor. 
Find, in terms of W and a, the tension of the rope and the reactions at the 
wall and the ground. (Assume that the weight of the ladder acts at its 
middle point.) 

3. The corner A of a square plate ABCD is held fixed by means of a 
smooth hinge which permits the plate to turn freely in its own plane. Four 
forces, each of magnitude P, act along the four sides in order. Find the 
single additional force which, applied at the center of the plate parallel to 
the side AB, will keep the plate in equilibrium. What is the corresponding 
reaction at the hinge? 

4. A door of weight W, height 2a, and width 26 is hinged at the top and 
bottom. If the reaction at the upper hinge has no vertical component, 
find the components of reaction at both hinges. (Assume that the weight 
of the door acts at its center.) 

6. A particle of weight W is suspended from a fixed 
point by a light string. A horizontal force H is applied to it, 
and the particle takes up a position of equilibrium with the 
string inclined to the vertical. If the string breaks when the 
tension in it roaches a value To, find the smallest value of // 
necessary to break the string. 

6. A heavy beam AB, 8 ft. long, rests horizontally on two 
supports, one at A and the other 3 ft. from B. If the greatest 
weight that can be hung from B without upsetting the beam is 
20 lb., find the weight of the beam. (Assume that the weight 
of the beam acts at its middle point.) 

7. Show that if a light cable passes round a pulley 
mounted on smooth bearings, the tensions in the portions on 
either side of the pulley are equal. Hence find the tension T 
in the cable for the pulley system shown, supporting a weight 
W, the pulleys and cable being supposed light and the distance 
between the upper and lower pulleys so great that the cables 
may be regarded as vertical. 

8. A force of magnitude P, acting up and along a smooth 
inclined plane, can support a weight W; when acting horizon- | W 
tally, it can support a weight w. Find a relation among P, W t 

and w, not involving the inclination of the plane. 

9. Four lamps each weighing 4 lb. are suspended across a road between 
posts 40 ft. apart by light cords attached at points B, C, D, E of a cord 



72 PLANE MECHANICS [Ex. II 

ABC DBF, whose ends A and F arc fixed at the same level to the posts. The 
cords supporting the lamps divide the horizontal distance between the posts 
into equal parts. C and D are 12 ft. below AF. Find the tension in CD. 

*10. A light rigid rod of length 26, terminated by heavy particles of 
weights w, W, is placed inside a smooth hemispherical bowl of radius a, 
which is fixed with its rim horizontal. If the particle of weight w rests 
just below the rim of the bowl, prove that 

wa* - W(2b* _ a 2). 

11. A system of forces acting on a rigid body consists of n forces acting 
along the n sides of a closed polygon taken in order. If the magnitudes of 
the forces are proportional to the lengths of the sides along which they act, 
show that the system reduces to a couple whose moment is proportional to 
the area enclosed by the polygon, a proper convention as regards the sign 
of this area being made. Give a simple example of such a system of forces 
which would keep a rigid body in equilibrium. 

12. Explain why in a motion picture the spokes of a rotating wheel some- 
times appear to be moving the wrong way. 

13. A force P of constant magnitude and fixed direction is applied to one 
end of an arm of length a, which can turn about the other end in a plane 
containing the direction of P. Find the total work done by the force as it 
pulls the arm into its own direction from a position perpendicular to it. 

14. Show that a field of force with components (X, Y) is conservative if, 
and only if, 



. 

dy dx' 

16. Find the potential energy of a particle attracted toward a fixed 
point by a force of magnitude k*/r n , r being the distance from the fixed 
point and fc, n any constants, 

*16. A light lever, in the form of a letter L with arms a and 6, is pivoted 
at the angle so that it can turn freely in a vertical plane. Weights W, w 
are suspended from the ends. Show that there are just two positions of 
equilibrium. 

*17. To a number of fixed points Ai, A- 2 , , A n , situated at equal 
intervals a on a straight line inclined at an angle to the horizontal, there 
are attached rods all of the same length a and weight w. The other ends 
of these rods, B\ t #2, , B*., are connected by rods of the same length a 
and weight w. The system hangs in a vertical plane, forming a set of 
squares, A\ and B 2 being connected by a light rigid rod. Find the reaction 
in this rod, assuming that all the joints are smooth and that the weight of 
each rod acts as its middle point. 

*18. A framework ABCD consists of four equal, light rods smoothly 
jointed together to form a square; it is suspended from a peg at A, and a 
weight W is attached to C, the framework being kept in shape by a light rod 
connecting B and D. Determine the thrust in this rod. 



Ex. II] METHODS OF PLANE STATICS 73 

19. A number of coplanar forces act on a rigid body. All the forces are 
turned in their plane through the same angle about their points of applica- 
tion, without change of magnitude. Show that their resultant turns 
through the angle about a fixed point in the body. (This point is called 
the astatic center.} 

20. Four forces of magnitudes 1, 3, 4, 6 act in order along the sides of a 
square ABCD of side a, the force of magnitude 1 acting along AB. Choosing 
as axes in the plane the lines AB and AD, find the equation of the line of 
action of the resultant force. Find also the position of the astatic center 
(see Exercise 19), if one force only is considered as acting through each 
corner of the square and the force of magnitude 1 acts at A. 



CHAPTER III 
APPLICATIONS IN PLANE STATICS 

In this chapter we shall be concerned chiefly with systems lying 
in a plane. However, mass centers and centers of gravity are 
here discussed for systems in space; the presence of a third 
coordinate causes no real complication. 

8.1. MASS CENTERS AND CENTERS OP GRAVITY 
Definition of mass center. 

Consider a system of n particles of masses mi, ra 2 , w n , 
situated at points Pi, P 2 , P. If the position vectors of 
these points relative to some assigned point are ri, r 2 , r n , 
we define the linear moment of the system with respect to that 
point to be the vector 



w t r t . 



The mass center of the system is defined to be that point with 
respect to which the linear moment vanishes. To show that 
this definition is significant, we have to prove two things: (i) a 
mass center exists; (ii) there is only one mass center. 

To establish the existence of a mass center, we take any 
point 0; let the position vectors of Pi, P 2 , P relative to 
be TI, r 2 , r n . Let C be the point such that 



(3.101) OC = 



Then the position vector of the point P< relative to C is 

it-OC, 
74 



SEC. 3.1] APPLICATIONS IN PLANE STATICS 75 

and so the linear moment of the system with respect to C is 

_ ^ n v 

m t (r t - OC) = m l r, - OC 



But this vanishes by (3.101), and therefore C is a mass center. 

To establish the uniqueness of the mass center, we assume 
that there are two mass centers C, C', relative to which the posi- 
tion vectors of the particles are TI, r 2 , r n and r{, r, r^ } 
respectively. Then, 

(3.102) 

But 

r. = rj + CC'; 

combined with (3.102), this leads to CC' = 0, so that C and C' 
coincide. 

Equation (3.101) gives the position vector of the mass center 
relative to an arbitrary origin 0; this position vector is the quotient 
of the linear moment by the total mass. It follows that the linear 
moment of a system is the same as that of a particle, having a 
mass equal to the total mass of the system, situated at its mass 
center. 

If the system consists of only two particles, with masses 
mi, w 2 , the definition shows that the mass center lies on the line 
joining them and divides it in the ratio m^:m\. 

For the calculation of mass centers, it is convenient to have 
(3.101) in scalar form; referred to any axes, the coordinates of 
the mass center are 



(3.103) x 



The numerators are, of course, the components of the linear 
1 moment with respect to the origin. 

It is important to note that when we move a system of particles 
rigidly (i.e., without changing mutual distances), the mass 
renter is carried along as if rigidly attached to the system. 



76 PLANE MECHANICS [SBC. 3.1 

This follows from (3.103). For let Oxyz be any set of axes and 
O'x'y'z' a new set of axes, such that the new position of the 
system relative to O'x'y'z' is the same as the old position relative 
to Oxyz; this means that x\ x l , y( y t , z( = z t . Then the 
coordinates of the new mass center relative to O'x'y'z' will be 
the same three numbers as the coordinates of the old mass 
center relative to Oxyz, and hence the new mass center occupies 
the same position relative to the system as the old one did. * 

Let us now consider a continuous distribution of matter instead 
of a system of particles. Viewing the continuous distribution as 
the limit of the discontinuous system, we aro led to associate a 
definite mass with any volume in the continuous distribution. 
Density is defined as mass per unit volume; by this we mean that 
the density p is 

(3.104) p = lim~?> 

whore Aw is the mass in the volume Av and the sign "lim" 
means "limit as Av contracts to a point. 7 ' In an infinitesimal 
volume dv the mass is 

(3.105) dm = pdv. 

In homogeneous bodies (with which we shall be chiefly concerned), 
p is a constant. If p varies from point to point in a body, the 
body is said to be heterogeneous. 

The definition given above for the linear moment of a dis- 
continuous system suggests that the linear moment of a con- 
tinuous system should be defined as 

JJ Jrp dx dy dz, 

where r is the position vector of a general point of the system and 
p the density at that point. This vector has components 

fffxp dx dy dz, J7/2/P dx dy dz, fff z P dx dy dz. 

The previous definition of mass center leads us to the statement 
that the mass center is that point for which, taken as origin, we 
have 

(3.106) ffjxp dx dy dz = 0, JJJ?/p dx dy dz = 0, 

J7/zp dx dy dz = 0. 



SEC. 3.1] f APPLICATIONS IN PLANE STATICS 77 

To find the mass center we may use (3.103), changed into 
continuous form. Thus, for any axes, the mass center haa 
coordinates 



r - dx dy dz _ Jf f yp dx dy dz 

X MJpdxdydz' J " J/Jp dz dy dz ' 

- _ 

"~ 



JJ JP da; ^2/ dz 

Consideration of a system of particles lying in or very close 
to a plane or surface leads to the idealized concept of a con- 
tinuous distribution of matter on a plane or surface; we intro- 
duce a quantity a called surface density, such that the mass of an 
infinitesimal area dS of the surface is <r dS. The mass center of a 
surface distribution has coordinates 



(3.108) 

v ' 



Similarly, we consider a continuous distribution of matter 
along a line or curve; we introduce a quantity X called the line 
density, such that the mass of an element ds is X ds. The mass 
center of a curvilinear distribution has coordinates 

. f?/X ds _ fzX r/s 

- - 



In (3.107), (3.108), and (3.109) the denominator in each caso 
represents the total mass of the system. 

In the case of uniform distributions of mass (i.e., distributions 
of constant volume density, surface density, or line density, as the 
case may be), the density factor comes outside the signs of inte- 
gration and so disappears by cancellation from (3.107), (3.108), 
and (3.109). 

Methods of symmetry and decomposition. 

In cases of symmetry, it is possible to locate the mass center 
(or, at any rate, limit its position) without any calculation. 

A system is said to have central symmetry with respect to a 
point if the system is left unchanged by reflection in the 
point 0. (By reflection we mean that a particle or element of 
mass m at A is replaced by a particle or element of mass m at 

B 9 where OB = OA.) For such a system, it is immediately 
seen that the mass center coincides with the center of sym- 



78 



PLANE MECHANICS 



[SEC. 3.1 



metry, because the linear moment about that point consists of 
contributions which cancel in pairs. 

A system has a plane of symmetry if the system is left unchanged 
by reflection in a plane. It is easily seen that in such cases the 
mass center lies in the plane of symmetry. 

A system has an axis of symmetry if the system is left unchanged 
by a rotation of arbitrary magnitude about the axis. It is 
not difficult to show that the mass center lies on the axis of 
symmetry. 

Thus, for example, it is evident that 

(i) The mass center of a solid sphere lies at its geometrical 
center, when the sphere is homogeneous or when the density 

depends only on the distance from 
the center. 

(ii) The mass center of a solid 
homogeneous hemisphere lies on the 
radius which is perpendicular to its 
plane face. 

(iii) The mass center of a plate 
in the form of an equilateral tri- 
angle (of uniform density and thick- 
ness) lies at the centroid. 

Sometimes we meet distributions 
of matter which may be decomposed 
into simple parts, the mass centers 
of which can be found. Such a sys- 
tem is shown in Fig. 28, the lines 
of decomposition being dotted. We shall now establish the fol- 
lowing principle of decomposition: If a system is decomposed into 
parts with masses M\, M*, M n and mass centers at the 



FIG. 28. A letter F is cut out 
of metal sheeting. The position 
of the mass center is required. 



points PI, P 2 , 



P, then the mass center of the complete 



system is at the mass center of the system of n particles of masses 
Mi, M z , - M n , situated at the points PI, P 2 , P w . 

We shall prove this principle for a system of particles, the 
proof for a continuous system being similar. Further, for 
simplicity we shall suppose that the system is decomposed into 
three parts, since the proof for n parts is similar. The proof 
rests on the fact that linear moments are additive; this is obvious 
from the definition of linear moment. Thus the linear moment 



SBC. 3.1] APPLICATIONS IN PLANE STATICS 79 

of the complete system is the sum of the linear moments of the 
three parts. But by (3.101) the linear moment of each part is 
the same as the linear moment of a particle situated at its mass 
center, having a mass equal to the mass of the part in question. 
Hence the linear moment of the complete system is equal to 
the sum of the linear moments of the three particles, and both 
vanish when they are calculated relative to the mass center of 
the complete system. This point is therefore the mass center of 
the three particles. 

In the case of the plate shown in Fig. 28, the reader should 
verify by this method that the mass center lies at x f-, 
y = ^. 

The principle of decomposition may also be expressed as 
follows : For the calculation of mass centers, any part of a system 
may be replaced by a representative particle, situated at the 
mass center of the part and having a mass equal to the mass of 
the part. 

In applying the method of decomposition, it is often convenient 
to decompose the system into infinitesimal portions. Generally 
the use of such a decomposition will require a process of integra- 
tion, but sometimes this can be avoided. Thus, if a triangular 
plate is decomposed into thin strips, the representative particles 
lie on the median of the triangle which bisects these strips. 
Hence the mass center lies on each of the medians; the mass 
center of a triangle is therefore at the ccntroid. 

By an extension of the same method, it is easily seen that the 
mass center of a solid tetrahedron lies at the point of inter- 
section of the lines joining the vertices to the ccntroids of the 
opposite faces. 

If we wish to find the mass center of a body with a hole in it, 
we can regard the body as a superposition of the complete body 
with no hole and a fictitious body of negative density (equal in 
absolute value to the density of the body) occupying the position 
of the hole. Thus, if a circular hole of radius 1 in. is punched 
from a circular disk of radius 4 in., the edge of the hole passing 
through the center of the disk, the mass center is that of a pair 
of particles with masses in the ratio 16 : 1 situated at the centers 
of the circles. Hence the mass center lies at a distance of -^r in- 
from the center of the larger circle. 



80 PLANE MECHANICS [Sao. 3.1 

Theorems of Pappus. 

Our knowledge of certain surface areas and volumes enables us 
to calculate some mass centers quickly by means of the theorems 
of Pappus, which state 

I. Let there be a uniform distribution of mass along a plane 
curve C, which does not cross a straight line L in the same plane. 
Let p be the distance of the mass center from L, I the length 
of C, and S the surface area generated by rotating C about L, 
to form a surface of revolution. Then 

(3.110) 2wpl - 8. 

II. Let there be a uniform distribution of mass on a region R 
of a plane. Let L be a line in the plane, not crossing R. Let p 
be the distance of the mass center from L, A the area of 72, and 
V the volume generated by rotating R about L, to form a solid 
of revolution. Then 

(3.111) 2*pA = V. 

To prove these theorems, we take axes Oxy, Ox lying along 
L in each case. Then, in the case of I, by (3.109) we have 

p = $y ds/l 

the integral being taken along C. But 
S = fay ds, 
and hence (3.110) follows. In the case of II, by (3.108) we have 

p = Jy dx dy/A, 
the integral being taken over R. But 

V = fay dx dy, 

and hence (3.111) follows. Thus the theorems of Pappus are 
established. 

As an example of the use of the first theorem, consider a wire 
bent into the form of a semicircle of radius a. We take for L the 
diameter joining the ends. Then 

(3.112) I = Tra, S = 47ra 2 , p = ~ = - 

Airl TT 

As an example of the use of the second theorem, consider a flat 
semicircular plate. We take for L the terminating diameter. 
Then 

(3.113) A = ^a*, V = fra', p = JL = g- 



SEC. 3.1] APPLICATIONS IN PLANE STATICS 81 

Mass centers found by integration. 

Though much labor may be saved by using the methods of 
symmetry and decomposition or the theorems of Pappus, it is 
evident from (3.107), (3.108), and (3.109) that when these 
methods fail we can fall back on direct integration. Usually a 
judicious mixture of the several methods will yield the result 
most rapidly. As illustrations, we shall calculate the mass 
centers of a wire bent to form a quadrant of a circle, a solid 
hemisphere, and a thin hemispherical shell. 

In terms of polar coordinates r, in its plane, the equation of a 
quadrant of a circle may be written r = a, with running from 
to -JTT. The length of an element is r dB] arid with x = r cos 0, 
y r sin 0, the Cartesian coordinates of the mass center are, 
by (3.109), 

2a 
x = j n a cos a av / / n a av 

(3.114) 



The mass center lies on the radius bisecting the arc at a distance 
2 -\/2 * a/ir from the center. The reader may compare (3.114) 
with (3.112) and consider how (3.114) might have been deduced 
from (3.112) without calculation. 

We may decompose a solid hemisphere into thin circular platen 
parallel to the plane face. The distance of the mass center from 
the plane face is thus 



'irr 2 dz/f a irr* dz, 

I Jz = * 



where r is the radius of the circular section at a distance z from 
the plane face. But r 2 a 2 z 2 , where a is the radius of the 
spherical surface. Hence, 

(3.115) I = |o. 

We may decompose a thin hemispherical shell into thin 
circular bands by means of planes drawn parallel to the open face. 
If is the angle between any radius and the radius perpendicular 
to the open face, the area of the band between and + d6 is 
2?ra 2 sin J0, where a is the radius of the shell. Hence the 
height of the mass center above the open face is 



82 PLANE MECHANICS [SEC. 3.1 

f ** a cos 6 - 2?ra 2 sin 6 dB 

(3.116) z = * ^ - = ia. 

/ 27ra 2 sin 0d0 

Historically, this result is famous; it was obtained by Archi- 
medes through comparison of the shell with a cylinder of the 
same radius, and length equal to the radius, containing the 
hemisphere and touching it along the edge of its open face. 
It is easy to show that two adjacent planes parallel to the open 
face intercept the same areas on the hemisphere and the cylinder. 
These two areas contribute the same linear moment, and so the 
mass centers of the hemisphere and the cylinder coincide; from 
this fact the result follows. 

Gravitation. 

A body falls to the ground unless it is held up by suitable 
forces. This is due to gravitational attraction between the body 
and the earth. Every body attracts every other body, and we 
accept as one of our hypotheses the following law: 

NEWTON'S LAW OF GRAVITATION. // two particles of masses 
mi, m% are at a distance r apart, each attracts the other with a 
gravitational force of magnitude 



where G is a universal constant, called the constant of gravitation. 
The forces act along the line joining the particles, in accordance 
with the law of action and reaction stated in Sec. 1.4. 

If we think of the particle of mass mi as fixed and that of mass 
m 2 as free to take up various positions, we recognize that the 
mass mi produces a field of force. It is usual to take m 2 = 1 for 
simplicity in discussing this field; then the magnitude of the force 
of attraction is Gmi/r 2 . 

If we take coordinates with origin at mi, the direction cosines 
of the line drawn from to any point A with coordinates x, y, z 
are x/r, y/r, z/r. Hence the components of force on unit mass 
at A are 



/'QIITN v i v i 

(3.117) X = -- -, Y = -- 3-, Z = 



the minus sign occurring since the force is directed from A 
toward 0. Now 



SEC. 3.1] 



APPLICATIONS IN PLANE STATICS 



83 



r 2 = z2 + 2/2 + 2 2 r|^ = s, - = -, 

d ' dz r 

therefore (3.117) may be written 

(3.118) X = ) Y -- ? Z 



d l 



where 
(3.119) 



7 = - 




FIG. 29. A spherical shell di- 
vided into thin rings for the cal- 
culation of the potential at A. 



This is the potential energy (cf. 2.419) of a particle of unit mass 
in the gravitational field of a particle of mass mi, or, briefly, the 
potential of the field. Thus the 
force of attraction is the gradient of 
the potential, with sign reverced. 
When a number of attracting 
particles are present, the resultant 
force of attraction is the vector 
sum of the individual forces of at- 
traction. This resultant force is 
equal to the negative of the gradi- 
ent of the total potential, i.e., the 
sum of the potentials due to the several particles. In calculat- 
ing the force of attraction due to a system of particles (or a 
continuous distribution of matter), it is often convenient to find 
the potential first. 

Let us consider a thin spherical shell of matter of radius a 
(Fig. 29). We wish to find tho potential at an external point A, 
at a distance r from the center 0. 

Let us draw cones with for vertex, OA for axis, and semi- 
vertical angles 0, 9 + dO. These cut off from the shell a ring of 
area 2ira 2 sin 6 dd. The elements of this ring are all at the same 
distance (R) from A, and so the potential due to the ring is 

-2irG<ra 2 sin Od6/R, 

where a is the mass per unit area of the shell. Expressing R hi 
terms of a, r, 6 and integrating over the shell, we find for the 
potential 

ra ian v - r 2wG<ra * s{nede 

(3.120) V- 



84 PLANE MECHANICS [Sac. 3.1 

the positive values of the square roots being understood. Since 
r > a, the last square root is r a, and so 

(3.121) . V = - 

/ i 

where M is the total mass of the shell. 

Thus we have the following result: The potential (and hence 
the force of attraction) of a thin spherical shell at any external point 
is the same as if the whole mass of the shell were concentrated at its 
center. 

If the point A lies inside the vshcll instead of outside, we proceed 
as before down to (3.120). But now a > r, and so the last 
square root is a r. Hence 

V = -4wO<ro, 

a constant. Thus, inside a thin spherical shell the potential is 
constant, and the force of attraction is zero. 

We can now discuss the gravitational field of the earth, sup- 
posing it to be composed of thin spherical shells, each of constant 
density. Each shell attracts as if its mass were concentrated 
at the center of the earth. Hence we have the following result: 
At a point A, outside the earth , the force of attraction is directed 
toward the center of the earth and is of magnitude 

(3-122) % 

where M is the mass of the earth and r the distance of A from the 
center of the earth. 

In particular, if r is the radius of the earth, (3.122) gives the 
force of attraction at the earth's surface. The constant G is very 
small (6.67 X 10~ 8 c.g.s. unit), and so gravitational forces are 
insignificant unless the masses involved are great. For this reason 
we usually neglect the mutual attractions of bodies on the earth's 
surface in comparison with the earth's attraction. 

Centers of gravity. 

We consider now a body near the earth's surface, the body 
being small in comparison with the earth's radius. (We have in 
mind a piece of laboratory apparatus or even a large engineering 
structure, but not anything which would be of appreciable size 
on a map of the world.) Throughout this body the direction and 



SEC. 3.1] APPLICATIONS IN PLANE STATICS 85 

magnitude of the earth's attraction are nearly constant. This 
leads us to the construction of the following model for the discus- 
sion of gravity near the earth's surface: The earth's surface (or the 
ground) is represented by a plane (the horizontal plane). The 
earth's attraction on a particle of mass m is of magnitude mg, where g 
is a constant; it is directed vertically downward (i.e., perpendicular 
to and toward the ground). The value of g is approximately 32 ft. 
sec.~ 2 , or 980 cm. scc.~~ 2 

We shall now show that there is just one point C, the center of 
gravity of a body, which satisfies the following conditions: 

(i) The potential energy of the body is equal to that of a single 
particle with mass equal to the total mass of the body, situated at C. 

(ii) The whole system of forces due to gravity is plane-equipollent 
(with respect to any vertical plane) to a single vertical force through C. 

Let us take axes Oxyz, Ox and Oz being horizontal and Oy 
vertical. Let us choose as standard position for each of the 
particles forming the body. Then a particle of mass ra t at the 
point (xj, yi, Zi) has by (2.120) potential energy n^gy*, and so 
the whole potential energy is (for n particles) 

(3.123) V = g 

Let the coordinates of C be x, y, z; condition (i) is equivalent to 

(3.124) V = Algy, 

where M is the total mass of the body. Henco, comparing the 
two expressions for V, we have 



n 

,1 



The condition of plane equipollencc with respect to the plane 
= demands that 



these being moments about Oz. Hence 



M 



86 PLANE MECHANICS [SEC. 3.2 

Similarly, 




Thus the center of gravity C exists, with coordinates 



7 7t It, 

2} mtX * 2) m #* 2) m ^ 

(3.125) 35 = r^ > y = Tjr ; % ~ jTr * 



We note, on referring to (3.103), that the center of gravity is in 
fact the same point as the mass center. 

The force Mg, directed downward through the center of 
gravity, is called the weight of the body. 

An accurate treatment of statics on the earth's surface is com- 
plicated by the earth's rotation about its axis and its motion 
round the sun. However, the effects due to these causes are 
very small, and we may neglect them without making serious 
physical errors. In fact, we get satisfactory results by treating 
the earth as a Newtonian frame of reference. Likewise, another 
simplification introduced above (the assumption that the earth is 
flat, with a uniform gravitational field) does not cause serious 
physical errors. So, if we do not wish to obtain results of 
extremely high physical accuracy, we may use the model described 
above; this is, in fact, the procedure throughout the rest of the 
chapter. 

The effects of the rotation of the earth are considered in Sec. 
5.3 and also in Sec. 13.5. It will be shown that, as far as statics 
is concerned, this introduces no real complication; it merely 
modifies the value of g. 

3.2. FRICTION 

In Sec. 2.4 we introduced the concept of a smooth surface; the 
essential property is that, at a smooth contact, the reaction is 
normal to the surface. We shall now discuss the reaction at a 
rough contact and state the laws of friction. 



SEC. 3.2] 



APPLICATIONS IN PLANE STATICS 



87 




Fio. 30. The reaction R at a 
rough contact resolved into the 
normal reaction (N) and the 
force of friction (F). 



Laws of static and kinetic friction. 

Let A and B (Fig. 30) be two bodies in contact. Let R be the 
reaction exerted by B on A. R can be resolved in a unique 
manner into the forces N and F, N 
lying along the normal at the point 
of contact and F lying in the plane of 
contact. N is called the normal re- 
action and F the force of friction. 
(At a smooth contact, F = 0.) 

On the basis of experiment, certain 
laws of friction are accepted. These 
are mathematical idealizations from 
the experimental results, and a high 
degree of accuracy in predictions 
based on these laws is not to be expected. 

LAW OF STATIC FRICTION. When two surfaces are in contact and 
no slipping takes place, the ratio F/N cannot exceed a number /u, the 
coefficient of static friction, which depends only on the nature 
of the surfaces. 

In statical problems the two bodies will be at rest, but the 
above statement is sufficiently general to cover the case where one 
body rolls on another. 

The acute angle X defined by 

(3.201) tan X = M 

is called the angle of friction. It is seen at once that the law of 
static friction 

(3.202) F ~ M 
implies 

(3.203) B <> X, 

where 6 is the inclination of the reaction R to the normal. Thus 
the direction of R must lie inside the cone of static friction, formed 
by drawing all lines inclined to the normal at an angle X. 

When one body slides on another, the behavior of the reaction 
is controlled by the law of kinetic friction. We shall state this 
law for the case where one body is at rest. 

LAW OF KINETIC FRICTION. When one surface slides on 
another which is at rest, the force of friction F on the former acts 



88 PLANE MECHANICS [Ssc. 3.2 

in the direction opposed to the direction of motion of the particle 
at the point of contact, and 

(3.204) - ', 

where it! is the coefficient of kinetic friction, which depends only on 
the nature of the surfaces. * 

If both surfaces are moving, the law has the same form except 
that the direction of the force of friction is opposed to the direc- 
tion of relative motion. 

The angle of kinetic friction X' is defined by 

(3.205) tan X' = /. 

As an experimental result, // is less than /z; jj, is always less than 
unity, t 

Problems in static friction often present considerable difficulty 
because the fundamental relation (3.202) is an inequality and, in 
mathematics, inequalities are usually more difficult to handle 
than equations. This difficulty may, however, be overcome by 
treating cases of limiting friction, for which 

(3.206) ~ = /i. 

When this relation holds, the system is on the point of slipping. 

Some problems on friction. 

Example 1. A light ladder is supported on a rough floor and leans against 
a smooth wall. How far up the ladder can a man climb without slipping taking 
place? 

In Fig. 31, AB is the ladder and C is the man (replaced by a particle). 
Only three forces act on the ladder: (i) the weight of the man (W}\ (ii) the 
reaction at the wall, this reaction being horizontal on account of the smooth- 
ness of the wall; (iii) the reaction of the ground. The lines of action of the 
first two meet at D. Hence the line of action of (iii) must pass through D, 
and hence the angle DBE t where BE is vertical, must not exceed the angle 
of friction X. Thus the highest position that the man can reach may be 
found as follows: Draw a line through B, making an angle X with BE; let it 
cut the horizontal through A at D; through D, draw a vertical; the point C 

* This quantity will be denoted by ju when there can be no confusion with 
the coefficient of static friction. 

t For further details regarding friction, see P. P. Ewald, Th. Poschl, and 
L. Prandtl, The Physics of Solids and Fluids (Blackie & Son, Ltd., Glasgow, 
1930), p. 67. 



SEC. 3.2] 



APPLICATIONS IN PLANE STATICS 



89 



where this line cuts the ladder is the required highest position. This 
method is called descriptive or graphical, because the result may be obtained 
by drawing to scale. 





F 



FIG. 31.- -The ladder 
problem for a smooth 
wal) (desciiptivo 
method). 



w 

Fio. 32.- The ladder 
problem for a smooth 
wall (analytical 
method) . 



Let us now discuss tho same problem analytically. Figure 32 shows the 
forces acting on the ladder. Lot be the inclination of the ladder to the 
vertical. The total vortical component must vanish; thus 

tf - W = 0. 
The total horizontal component must vanish; thus 

N' - F = 0. 
The total moment about B must vanish; thus 

W BC sin a N' - AB cos 0. 



Hence 



Thus, by (3.202), 



F = N' - W 



BC 
'AB 



tan a, 



N 
F 

N 



w, 

BC 

"AB 

BC 
TB 



tan a. 



. 

tan a <> 



The highest point C attainable is given by 
(3.207) BC ABn cot a. 

The analytical method appears more complicated than the descriptive, 
but it has the advantage of being more systematic. Moreover, since the 



90 



PLANE MECHANICS 



[3EC. 3.2 



three conditions of equilibrium give all possible information, the solution of 
the problem is reduced to algebra as soon as they are written down. 

It might be thought that in drawing the arrow for the force of friction to 

the left in Fig. 32, we were antici- 
pating the result. This is not ac- 
tually the case. When we draw 
an arrow in connection with a com- 
ponent of a force, we are simply 
indicating the sense in which this 
component is considered positive. 
Had we drawn the arrow to the 
right in Fig. 32, we should have ob- 
tained equations as above, but with 
the sign of F reversed. The final 
physical result would have been the 



However, since positive quanti- 
ties are easier to think of than nega- 
tive quantities, it is advisable 
whenever possible to draw the 
arrows in the senses in which the 
forces really act. Thus, in the case 
of N, we draw the arrow upward. 
As for friction, it is generally found 




B 
W1 

FIG. 33. The ladder problem for 
rough wall (descriptive method). 



that the force of friction acts in the direction opposed to the motion which 
would take place in its absence. That is why the arrow for F in Fig. 32 was 

drawn to the left. 

Example 2. The preceding problem 
modified by supposing both wall and 
floor to be rough, with the same coefficient 
of friction /*. 

Consider the cones of friction at A 
and B. They will cut the plane of the 
paper in four lines as shown in Fig. 33, 
these four lines giving the quadrilat- 
eral FGHJ. Draw the vertical 
through C, the position of the man, 
and let this vertical cut the sides of the 
quadrilateral at K, L. Let M be any 
point on the segment KL. Now the 
weight W may be resolved into forces 
along MA, MB, and hence W can be 
balanced by forces along AM, BM. 




B 

FIG. 34. The ladder problem for a 
rough wall (analytical method). 



Since these lines lie inside the cones of friction, the law of friction is satisfied. 
We have here a case of statical indeterminacy (cf . Sec. 2.5) : provided that 
the vertical through C cuts the quadrilateral FGHJ, the ladder will be in 
equilibrium, but we cannot tell precisely what the reactions of the wall and 
floor will be. 



SEC. 3.2] 



APPLICATIONS IN PLANE STATICS 



91 



Now let us ask: How far can the man go up the ladder before slipping 
takes place? Obviously, he can climb until the vertical through his position 
passes through the point J. When he passes that position, it will no longer 
be possible to find reactions satisfying the conditions of equilibrium and the 
law of friction. 

The question may also be treated analytically. Consider the man slowly 
climbing the ladder. If the ladder slips at all, just at the point of slipping 
the reactions at both contacts must correspond to limiting friction. Thus, 
at the point of slipping, the force system is as shown in Fig. 34, with F = pN. 
F' pN'. Taking vertical and horizontal components and moments about 
B, we have the three equations 



W BC sin a - 



+ N - W - 0, 

N' - N = 0, 

- AB sin o - AT' 



AB cos a - 0. 



These three equations determine N, N r , BC: we find 

W , W BC .._ 

1 +// 



(3.208) N 



N' 



AB 



). 



w 



Fio. 36. A heavy block pushed by a horizontal force P. 

Example 3. A block rests on a rough horizontal floor and is pushed by a 
gradually increasing horizontal force. Will the block slide t or will it topple 
over an edge? 

Let the thickness of the block be 2a, its weight W, and the coefficient of 
static friction /*. Let the horizontal force P be applied at a height h above 
the floor. The first question is: Given W and P as shown in Fig. 35, can 
there be a system of reactions exerted by the ground, satisfying simultane- 
ously the law of friction and the conditions of equilibrium for the block? 
Any such system of reactions will be plane-equipollent to forces X, Y at A as 
shown, together with a couple N. If equilibrium exists, it is clear that the 
following conditions are demanded by the law of friction and the fact that 
the floor cannot pull the block downward : 



(3.209) 



Y >. 0, N 2> 0. 



92 



PLANE MECHANICS 



[SEC. 3.3 



Taking horizontal and vertical components and moments about A, we 
have 

X = P, F - W, N - aTF - AP, 

and so (3.209) give 

(3.210) P <Z nW, P ^ ^ 

Starting with a small value of P, these inequalities are both satisfied; 
but as P is increased, one or other will be violated, and then equilibrium will 
cease. If 

(3.211) ft < *> 

the first inequality of (3.210) will be broken first. At the instant when 
P = nW, we have 

X-pY, N > 0. 

This is a state of limiting friction; and so, if (3.211) holds, equilibrium of the 
block will be broken by sliding along the plane. On the other hand, if 

(3.212) M > , 

then the second inequality of (3.210) will be violated first. At the instant 
when P = aW/h, we have 

X < pY, Y > 0, N 0. 

The friction is not limiting, and so slipping cannot take place. But any 
further increase in P will cause violation of the last inequality of (3.209). 
Hence we conclude that, if (3.212) holds, equilibrium will be broken by 
the block turning over the edge A. 

The result is in agreement with common experience: the smaller we 
make h, the more likely is sliding to occur. 

3.3. THIN BEAMS 
Tension, shearing force, and bending moment. 

Let us consider a straight beam of uniform section (Fig. 36) 
and a plane P parallel to its length. P may be regarded as the 

S 



A 




' -*xM 


/' 


1 \ 


/ 

B 


A 

/ 


lo 


1 }T 


j * 1 


!/ 


/ 



FIG. 36. Reactions across a section of a beam. 

plane of the paper. External forces, parallel to P, act on the 
beam. (These forces are not shown. They may consist of 
the weight of the beam or loads placed on it.) Let a cross section 




SEC. 3.3] APPLICATIONS IN PLANE STATICS 93 

be drawn through a point O, perpendicular to the length of the 
beam. Let us take as our "system" the portion of the beam 
extending from the end A up to this section. The external forces 
acting on this system will consist of 

(i) the external forces already mentioned, acting on this por- 
tion of the beam, 

(ii) the reactions exerted across the section by the particles 
in the portion of the beam extending from the section to the end 
A (J5 B 

FIG. 37. A thin beam?" 

B. These reactions are internal forces as far as the whole beam 
is concerned, but thejr are external forces for the system at 
present under consideration. 

Let us take P as the fundamental plane. The reactions across 
the section are plane-equipollent to a force acting at 0, together 
with a couple M. The force may be resolved into components 
T, S along the beam and per- 
pendicular to its length, re- 
spectively. We define the A 
following terms: 

T = tension, 
S = shearing force, 
M = bending moment. 

We shall confine our atten- (6) 

tion to thin beams. The thin Fio. 38. (a) Reactions exerted on 

beam is a mathematical ideal- $%$ (&) RcaPtlons oxorted on 
ization, in which the cross sec- 
tion is reduced to a point and the beam to a straight line. 

Figure 37 shows a thin beam AB; C is any point of it. To 
draw the reactions on AC across the section at C without con- 
fusion, we delete the line CB as in Fig. 38o. Figure 386 shows 
the reactions on CB; these have the same magnitudes as, but 
opposite senses to, those shown in Fig. 38a, on account of the 
law of action and reaction. 

Let us take an origin on the beam, the rr-axis along the beam 
and the y-axis perpendicular to it. Consider a small length of 
the beam extending from x to x + dx (Fig. 39). Let T, S, M 
be the values of tension, shearing force, and bending moment at 
x, and T + dT, S + dS, M + dM the values at x + dx. To 
allow for gravity or other continuous external loading, we shall 



94 



PLANE MECHANICS 



[SEC. 3.3 



add a force with components -X" dx, Y dx (not shown in Fig. 39) 
acting at the middle point of the portion x, x + dx. By taking 
components and moments about the point x and neglecting 
y 



S-hdS 




FIG. 39. Reactions on the ends of a small element of a beam. 

infinitesimals of the second order, we have, as conditions of 
equilibrium for the small length of the beam, 

dT + X dx = 0, dS + Y dx 0, dM + S dx = 0. 
Thus 

(3.301) --Z, --Y, --* 

v ' ' ' 



dx 



dx 





dx 



B 



These are the general differential equations for the equilibrium of 

thin beams. But in statically 
determinate cases we can obtain 
all required information regard- 
ing internal reactions without us- 
ing these equations, or rather 
by using them in integrated 

form 



w 

a. 40. A light beam loaded at its 
middle point. 



Statically determinate problems. 

We shall illustrate the method by the solution of a problem. 
A light beam A B of length 2a is hinged at A and supported on a 
smooth horizontal plane at B (Fig. 40). A load W is placed at 
the middle point C. Find the bending moment and shearing 
force along the beam. 

First, by application of the conditions of equilibrium (2.306) 
to the whole beam, we find the reactions on the beam at A and B. 



SEC. 3.3] 



APPLICATIONS IN PLANE STATICS 



95 



These are each of magnitude %W , directed upward. Let us take 
our origin at A and the z-axis along the beam. Consider the 






r tsf 

|< - X >l > ip 

A D 

Fio. 41. External 
forces on a portion of 
the beam shown in Fig. 
40 (AD < AC). 



D 



W 

Fio. 42. External forces on a 
portion of the beam shown in 
Fig. 40 (AD > AC). 



>T 



(3.302) 



portion of the beam AD t where D lies in A (7; let AD = x (Fig. 
41). From the equilibrium of AD, we have 
(T = 0, 8 = -iTF, 
\M = -xS = &W, (x < a). 
These give the shearing force and bending moment for any point 
m AC; there is no tension. Since S is negative, the shearing 
force actually acts in the downward direction. 

Now take D in CB (Fig. 42). Instead of (3.302), we have 



- xS = (a - 
The shearing force is now positive. 
(3.301) is satisfied by (3.302) 
and (3.303). 

The graphs of S and M along 
the beam are shown in Fig. 43. A 



We note that the last of 




M 



The Euler-Bernoulli theory of 
thin elastic beams. 



-w 


c 



8 



FIQ. 43. Graphs of shearing force 
(S) and bending moment (M) along 
the beam shown in Fig. 40. 



If a straight beam rests on 
three supports, the problem of 
finding the reactions due to the supports is statically indeter- 
minate (cf. Sec. 2.5), and we cannot find the shearing force and 
bending moment by elementary statical principles. But this 
indeterminacy disappears when we take into consideration the 
elasticity of the beam. Although straight initially, an elastic 
beam will stretch and bend under the influence of forces. We 



96 PLANE MECHANICS [SEC. 3.3 

suppose the stretching and bending to be very small and accept 
the law of Hooke for stretching and the law of Eulcr and Ber- 
noulli for bending.* 

HOOKK'S LAW. When a beam is slightly stretched, 

(3.304) T = k'e, 

where c is the extension (increase in length per unit length) and 
k r a constant for the beam. (Actually k r = EA, where E is 
Young's modulus for the material and A the area of the cross 
section.) 

THE EULER-BERNOTJLLI LAW. When a beam is slightly bent, 
the bending moment is connected with the curvature by the 
relation 

(3.305) M = -> 

P 

where p is the radius of curvature and k a constant for the beam. 
(Actually k = El, where E is Young's modulus and 7 the 
"moment of inertia" of the cross section about an axis through 
its mean center perpendicular to the plane of the couple M.) 

When the beam is approximately straight and the axes as in 
Fig. 39, 

p = ~dx* approximately, 

and (3.305) may be written 

(3.306) M = k g- 

Let us refer to Fig. 39 and to the equations (3.301). We shall 
suppose that the beam is subject to a force w per unit length 
in the negative sense of the 2/-axis, due either to its own weight 
or to a load placed on it. Then X = 0, Y = w, and (3.301) 
read 

(3.307) f = 0, f = , d -f=-S. 

^ ' dx ' dx ' dx 

We see that the tension T is constant. Elimination of M and S 
from (3.306) and (3.307) gives 

(3.308) *g -u,. 

* The law of Euler and Bernoulli follows from that of Hooke; the proof 
belongs to the theory of elasticity. 



SEC. 3.3J 



APPLICATIONS IN PLANE STATICS 



97 



This is the fundamental differential equation in the theory of 
thin elastic beams. If it is solved, the bending moment and 
shearing force are given by 

d*y s = _ dM 
dx v ~ dx 



(3.309) 



M 



S+AS 



AM+AM 



It must be realized that the differential equation (3.308) holds 
only between isolated loads or supports. To deal with these a 
special treatment is necessary. 
Figure 44 shows an clement Ax 
of a thin beam with an isolated 
load W suspended from its 
middle point P. (The case 
of a support is covered by 
making W negative.) 

The element is in equili- 
brium under four forces and 
two couples: the continuous 
load on Ax (not shown), the 
isolated load W, the shearing 
force S + AS, and the bend- 
ing moment M + AM on the 
right, and the shearing force 
S and the bending moment M 
on the left, positive senses 
being as indicated. 

If Ax tends to zero, the continuous load tends to zero and so 
does the moment of this load about P. Hence the conditions 
of equilibrium give, in the limit, AS = W, and (taking moments 
about P) AM = 0. 

This means that the bending moment M is continuous across 
an isolated load or support, but the shearing force S changes 
abruptly. In terms of y and its derivatives (since the beam is 
not broken at the isolated load or support) we have continuity 
in y, dy/dxy d^y/dx*, but discontinuity in d*y/dx*. 

Example. A uniform heavy beam OP of length 2a and weight W is hinged 
at O and rests on two smooth supports, one at P and the other at its middle point 
Q. Find the reactions on the supports, if 0, P, Q are all at the same height. 

We shall take the origin of coordinates at O, the re-axis horizontal, and the 
y-axis directed vertically upward. Integration of (3.308) along OQ gives 



FIG. 44.- -Element of beam containing 
ibolatcd load. 



(3.310) 



ky - - faux* -f Ax* + Bx t 



(OQ) 



98 PLANE MECHANICS [Sac. 3.4 

where A, B are constants of integration; two other constants of integration 
have been put equal to zero on account of the vanishing of y and d*y/dx* at O 
(There can be no bending moment at a hinge or free end ) Similarly, we 
have along QP 

(3.311) ky = -ftw(x - 2a)< + A'(x - 2a) + B'(x - 2a), (QP} 

where A', B' are constants of integration. In these two equations, we have 
four unknown constants; they are to be found from the conditions that 
y = at Q, while dy/dx and d*y/dx* are continuous there. Thus, we have 
the four equations 



Aa 3 + Ba fWa* = 0, 
A 'a 8 + B'a + fawa* = 0, 

3Aa 2 + B - \wa* - 3A'a 2 + B' 
6Aa - Jfl = 6 A 'a 
We find 

A. -A' = ^wa, # = B' = 

and substitution in (3.310) and (3.311) gives the equations of the two por- 
tions of the beam 



~ ar) 4 4. 
In OQ the bending moment is 



Its maximum value occurs at x = |<z. The portion OQ is a system in equi- 
librium ; hence, taking moments about Q, we have for the reaction JRo at O 

Rod - M Q + $wa z - |u>a 2 - &Wa. 

When one reaction has been found, the others follow from the usual statical 
methods. Hence 



(3.313) Ro = &W, R Q - i$JK, R P = 

3.4. FLEXIBLE CABLES 

A flexible cable differs from a stiff rod in the ease with which it 
can be bent into a curve. The bending moment per unit curva- 
ture is much less for the cable. In mechanics, we idealize this 
property and understand by a flexible cable a material curve such 
that there can be no bending moment across any section. By 
considering the equilibrium of a small portion of the cable, it is 
easily seen that the shearing force must also vanish. Hence the 
only surviving component of the reaction across a section of a 
flexible cable is a tension 2 7 , which acts along the tangent to the 
curve in which the cable lies. 

We use the word "cable " exclusively, but it is to be understood 
that the practical applications cover chains, ropes, strings, and 
threads. The theoretical predictions will agree well with the 



SEC. 3.4] 



APPLICATIONS IN PLANE STATICS 



99 




Wdft 



results of experiments conducted on cables in which the bending 
moments are small. 

General formulas for all flexible cables hanging freely. 

Let us consider a flexible cable hanging under the influence 
of its own weight, and perhaps additional continuous vertical 
loads attached to it. For the 
present, we shall not make any 
special assumptions regarding the 
nature of the cable or the load. 

We pass over the trivial case 
in which the cable hangs from 
one end in a vertical line. When 
suspended from two points, it 
hangs in a vertical plane. Let 
Oxy be axes in this plane, Ox being 
horizontal and Oy directed verti- 
cally upward (Fig. 45). Let A 
be a point on the cable with coordinates (.r, y), and B an adjacent 
point with coordinates (x + dx, y + dy). Let ds be the infini- 
tesimal length of AB, and let w ds be the total load on AB, 
including the weight of the cable. The portion AB is a system 
in equilibrium under the action of the tensions at its ends and 
the load. Let B be the inclination of the tangent at A to the 
horizontal. Then dx/ds = cos 0, dy/ds = sin 0; and so, taking 
horizontal and vertical components, we have 



0. 



By the first of these equations, the horizontal component of the 
tension is constant. The second equation may be written 



FIG. 45. Forces acting on an ele- 
ment of a hanging cable. 



(3.401) 

If // is the constant horizontal component of tension, we have 

(3.402) rg-H; 

substitution in (3.401) gives 

<n> sSD-r 



100 



PLANE MECHANICS 



[SEC. 3.4 



This is a differential equation satisfied by the curve in which the 
cable hangs. When this equation has been solved, the tension 
may be found from (3.402). 

The suspension bridge. 

Let us now suppose that a weightless cable supports a load 

uniformly distributed on a hori- 
zontal line; for the load on a hori- 
zontal length dx, we write w dor. 
This approximates to the condi- 
; tion of a cable of a suspension 
bridge (Fig. 46), the load consist- 
ing of the roadway AB, of weight 
W Q per unit length. 

With the notation used above, we have w ds = WQ dx, and so 

dx 





A B 

FIG. 46. Suspension bridge. 



thus (3.403) reads 



ds \dx 



77 5? 



dx* 



H 



If the origin is chosen at the lowest point of the cable, so that 
y = dy/dx = when x = 0, we obtain as the equation of the 
cable 

(3.404) y = i^ 



This is a parabola. The tension in the cable is given by (3.402). 
Since 



(3.405) 
we have 

(3.406) 






T = H 



The common catenary. 

We shall now consider a uniform cable hanging freely under its 
own weight, w per unit length. The fundamental equation is 



SBC. 3.4] 



APPLICATIONS IN PLANE STATICS 



101 



(3.403), in which w is now a constant. We write it in the form 

d*y _ w ds 

dx* ~ H fa' 
or, by (3.405), 



(3.407) - 

Introducing a variable z defined by 

(3.408) sinh z = ~|, 
we reduce (3.407) to 



dx 



w 
H' 



and so 




47. Tho common 
catenary. 



where A is a constant of integration. 

Choosing the origin O at the lowest 

point of the cable (Fig. 47), we have y = dy/dx = fora; = 0, 

and hence z = for x = 0. Thus A = 0, and (3.408) reads 



(3.409) 

Hence 

(3.410) 



dy . t wx 
= smh 



H ( 
y = I 

J w \ 



, wx . 
cosh -jf 1 



when account is taken of the conditions at 0. This curve is 
called the common catenary; the lowest point is called its 
vertex. 

It is customary to define the parameter c of the catenary by 

(3.411) c - ; 
then (3.410) reads 

(3.412) y = c ( cosh ^ - 1 

To find the tension from (3.402), we note that from (3.405) 
and (3.409) 



(3.413) 



102 



PLANE MECHANICS 



[SBC. 3.4 



and so 
(3.414) 



T = H ~ = H cosh - = H + wy. 
ax c 



So far we have concentrated our attention on two things, 
the curve in which the cable hangs and the tension at any 
point in it. These have been found in (3.412) and (3.414). 
But other problems suggest themselves, and we need other 
formulas to solve them. Such problems may involve the length 
of the cable and the inclination of its tangent to the horizontal. 
Let us denote the length by s (measured from the vertex to a 
general point) and the inclination by 0} there are then five 
variables involved in the theory of the cable, 

x, y, T, 5, 6. 

Any one of these variables is expressible in terms of any other, 
and it is an interesting exercise to prepare a table of five rows and 
columns showing all the twenty expressions. We shall note here 
only the expressions giving s in terms of x, y, and 0, as follows: 



(3.415) 
(3.416) 



s = c sinh -, 
c 



s 2 = y 2 + 2yc t 
c tan 8. 



These equations are easy to obtain from (3.413), combined with 
(3.412) ; to get (3.416), we use the fact that dy/dx = tan 0. The 
equation (3.416) is the intrinsic equa- 
tion of the catenary. 

Examples. Problems connected with 
freely hanging cables usually involve the so- 
lution of a transcendental equation. As 
illustrations, two problems will be considered. 
These problems may be stated briefly as 
follows : 

(i) Given the span and length, to find the 
maximum tension. 

(ii) Given the length and sag, to find the span. 




FIG. 48. A hanging 



A cable, of weight w per unit length and length 2Z, hangs from two points 
A and B, at the same height and at a distance 2a apart (Fig. 48). We wish 
to find the maximum tension in the cable. 

It is clear from (3.414) that the maximum tension occurs at A and B, and 
the value is 



(3.417) 



nM = H cosh - = we cosh 
c c 



SEC. 3.4] APPLICATIONS IN PLANE STATICS 103 

Here, as in most problems on the catenary, the solution depends on finding 
the parameter c. Applying the first of (3.415) at the point B, we have 

(3.418) I c sinh i 

This is an equation to determine c in terms of a and I; it may be written 

(3.419) B -^/) - L 



If tables of (sinh X)/X are available, the numerical value of a/c may be 
obtained at once.* The solution of the problem is given by (3.417) on 
inserting the value for c, found from (3.419). 

If the ratio I /a is nearly unity, i.e., if the cable is only a little longer than 
the span, the solution of (3.419) for a/c is small, because 

, sinh X .. 
hm y = 1. 

In fact, the parameter c is large. Then we can obtain an approximate solu- 
tion of (3.419) without recourse to numerical tables. Retaining only the 
first two terms of the expansion for sinh a/c, we have 

(3.420) 




Since c is large, T mt * as given by (3.417) is large; it is approximately equal 
to H, where 

(3.421) // = we - wa ^^~ a y 

The second problem arises when the distance between two points A and B 
at the same height is measured by a measuring tape which sags under its 
own weight. With the notation of Fig. 48, we arc given h, I; we wish to 
find a. 

Applying the second of (3.415) at the point B, we have 

(3.422) c = ~ - - 

The answer to the problem is given by (3.418). This is a quadratic equa- 
tion for e a/c , and the positive root gives t 



* J. W. Campbell, Numerical Tables of Hyperbolic and Other Functions 
(Houghton Mifflin Company, Boston, 1929), p. 30. These tables were 
prepared with the solution of catenary problems in mind. 

t Throughout this book "log" moans the natural logarithm, that is, log.. 



104 PLANE MECHANICS [Sac. 3.4 

If the ratio h/l is small, we have approximately 
(3.424) lc 

and hence 
(3.425) 
with an error of the order of 

Cables in contact with smooth curves. 

So far the cables considered have been unconstrained. Let us 
now consider the case of a cable lying against a smooth surface, or, 
as we may say in two-dimensional language, against a smooth 
curve. Gravity will be neglected. 

Figure 49 shows a small portion AB of a cable lying in equilib- 
rium in contact with a smooth curve. Let be any assigned 

point on the cable and s the length 
of the cable between and A. 
Let the length AB be ds, and let 
the inclinations to some fixed di- 
rection of the tangents to the cable 
at A and B be and + d9. The 
element A B is in equilibrium under 
three forces, namely, the tension T 
at A, the tension T + dT at B, and 
a normal reaction due to the curve. 
This last may be written N ds and 
may be supposed to act along the 
normal at A. Resolving forces along the tangent and normal at 
A, we obtain from the conditions of equilibrium 



Nds 




FIG. 49. Forces on an element 
of light cable in contact with a 
smooth curve. 



(3.426) 



dT = 0, N ds = T dO. 



Hence, the tension is constant along a light cable in contact with a 
smooth curve. Also, since ds/dO = p, the radius of curvature, 
we have 



(3.427) 



T 
N = - 



An example of the significance of this last formula occurs in 
tying up a parcel : the sharper the edge of the parcel, the smaller p 
and hence the greater the tendency of the string to bite into the 
parcel. 



SBC. 3.4] APPLICATIONS IN PLANE STATICS 105 

Cables in contact with rough curves. 

Let us now suppose that the curve shown in Fig. 49 is rough 
and that the cable is just on the point of slipping in the direction 
AB. In addition to the forces already considered, there is now 
a force of friction F ds on the element, acting along the tangent 
at A and opposing motion. The conditions of equilibrium are 
now 

(3.428) dT = F ds, N ds = T dO. 

But F = pN, where /x is the coefficient of friction. Hence 

(3.429) g.^, tf-rg 
and so 

(3.430) = ,T. 

Integration gives 

(3.431) T = To &', 

where T is a constant of integration. 

The rapid increase of the exponential with increasing is of 
great practical importance. As a numerical example, consider a 
rope wrapped twice around a post, for which the coefficient of 
friction is . Then 

T = T Q C** = To e 2 *, 

where T , T are the tensions in the rope where it meets and 
leaves the post, slipping being about to occur in the direction 
of T. We have 

?J> = c -2ir = 0.0019. 

If T = 2000 lb., To = 3.8 Ib. Thus, a load of one ton can be 
sustained by application of a force of less than 4 lb.; and, of 
course, a much greater load might be sustained if the rope were 
wrapped more often round the post. This principle is used in 
holding ships by ropes passed round mooring posts and in hoists 
in which a rope is passed round a revolving drum, the end being 
held in the hand. 



106 PLANE MECHANICS [SEC. 3.5 

3.6. FRAMES 
Just-rigid frames. 

Figure 50 shows a simple frame or truss, as used in bridges. 
It consists of steel girders riveted together at the joints. For 
mathematical discussion we simplify the system as follows: 
(i) the girders are treated as light rigid bars, (ii) the joints are 

C D 




FIG. 50. A just-rigid frame with loads applied at E and F. 

supposed to be smoothly working hinges, each bar being capable 
of rotation about the joints on it without any resisting couple. 
We shall discuss only frames with joints lying in a plane, and 
we shall not consider displacements out of that plane. 

In Fig. 50, we suppose the joint A fixed and the joint B con- 
strained to slide on a horizontal line. Inspection shows that the 
whole frame is fixed by these conditions. In fact, the frame is a 
rigid body and is fixed when one of its points is fixed and another 
of its points constrained to move on a line. If one bar, for 
example CD, were removed, the frame would cease to be a rigid 
body. Hence it is called just-rigid. The following is the general 
definition : A frame is just-rigid when the removal of any one of its 
bars destroys its rigidity, 

If an additional bar is inserted in a just-rigid frame, it becomes 
over-rigid. We shall deal only with just-rigid frames. 

We shall now show that a just-rigid frame with j joints has 
2j 3 bars. Taking any axes in the plane of the frame, we 
denote the coordinates of the joints by (xi, 2/1), (x 2 , 2/2), 
(xj, 7/7); there are 2j coordinates altogether. If the first two 
joints are connected by a bar of length l y their coordinates must 
satisfy 

(xi - * 2 ) 2 + (2/1 - 2/ 2 ) 2 l\ 

Thus if there are 6 bars, the 2j coordinates are subjected to 6 
relations of this type. If we fix one joint and constrain another 



SEC. 3.5] APPLICATIONS IN PLANE STATICS 107 

joint to move on a line, we impose 3 more conditions. If the 
frame is just-rigid, these 6 + 3 conditions suffice to fix the whole 
frame, i.e., to determine the 2j coordinates of the joints. Hence 
b + 3 = 2jj which gives the stated result: 

(3.501) 6 = 2j - 3. 

If 6 < 2j - 3, the frame is not rigid. r FlG> 51 .__Frame 

The smallest number of joints possible in a with three joints 

just-rigid frame is j = 3. Then the number and three bar9 ' 

of bars is 2j 3 = 3. In this case, we have a triangular frame 

(Fig. 51). 
Now take j = 4; then the number of bars is 2j 3 = 5. 

Examples are shown in Fig. 52. (When two bars cross in a 






FIG. 52. Frames \uth four joints and five bars. 

diagram, without indication of a joint, they are supposed capable 
of free motion past one another.) 

If j = 5, the number of bars is 2j 3 = 7. Examples 
are shown in Fig. 53. 




FIG. 53. Frames with five joints and seven bais. 

A simple way to build up a just-rigid frame is to start with a 
triangle and add two bars at a time. Since, in each operation, 
we add one joint and two bars, after p operations we have 3 + p 
joints and 3 + 2p bars; the identity 

3 + 2p 2(3 + p) - 3 

shows that the condition for a just-rigid frame is satisfied. 
However, all just-rigid frames cannot be constructed in this way. 



108 PLANE MECHANICS [Sac. 3.5 

Stresses in bars. 

Suppose that a just-rigid frame is fixed by external constraints 
so that it cannot move. (The normal plan is to fix one joint and 
constrain another joint to move on a line, as in Fig. 50.) Now 
let external forces, or loads, be applied to some or all of the joints. 
Each bar is in equilibrium under two forces, the reactions at its 
ends. These two forces must be equal in magnitude and act in 
opposite senses along the bar. If the forces act away from one 
another (so that the bar tends to be torn in two), the bar is said to 
be in tension; if the forces act toward one another (so that the bar 
tends to buckle), the bar is said to be in thrust. The word stress 



Tension Thiust 

Fio. 54. Reactions exerted by a bar on tho joints at its ends. 

is used to cover both cases. A plus sign is associated with 
tension and a minus sign with thrust. Thus, if we say that the 
stress in a bar is +3 tons, we mean that there is a tension of 3 tons 
in it; if the stress is 5 tons, we mean that there is a thrust of 
5 tons. In Fig. 54 the arrows indicate forces exerted on the joints 
by the bars. The forces exerted on the bars by the joints art 
in the reverse directions. 

For a frame in equilibrium, two problems arise: 
(i) to determine the external reactions at the supported joints; 
(ii) to determine the stresses in the bars. 

The first problem is elementary. It is a question of the equi- 
librium of a system, as discussed in Sees. 2.3 and 2.4 and sum- 
marized in Sec. 2.C. It is with the second problem that we arc 
concerned. 

Method of joints. 

The following argument is general, but the reader may con- 
sider the frame shown in Fig. 50 as an example, the loads being 
indicated by arrows at the joints E and F. The loads arc given, 
and the stresses are to be found. Each joint may be considered 
as a particle in equilibrium, under tho action of a load (if any) 
and the reactions of the bars meeting there. (Since the joint is 
the system considered, this method is called the method of joints.) 
As the forces lie in a plane, there arc two equations of equilibrium 
for each joint, and thus 2j equations in all if tho number of joints 
is j. These equations involve 3 unknown components of external 



SEC. 3.5] 



APPLICATIONS IN PLANE STATICS 



109 



reactions at the supports and a number of unknown stresses 
equal to the number of bars, i.e., 2j 3. Thus the total 
number of unknowns is 2j, and we have 2j linear equations to 
find them. Thus, in a just-rigid frame the problem of finding the 
external reactions at the supports and the stresses in the bars is a 
determinate problem, involving the solution of a number of simul- 
taneous linear equations equal to twice the number of joints. 

If the frame were over-rigid, the number of unknowns would 
exceed the number of equations, and the problem would be 
indeterminate. We should have to consider the elastic properties 
of the bars. 




FIG. 55a. A frame with 14 joints, supporting a load at M. 



The problem of the just-rigid frame having been thus reduced 
to the solution of simultaneous linear equations, it might be 
thought that nothing remained to be said. However, the 
system of equations obtained in the manner described above 
may be very involved, and much labor may be avoided by 
modifying the method. This is particularly true if we only 
require the stresses in certain bars. But the reader should 
realize that these are only laborsaving devices. If he cannot 
discover the particular device suited to a certain problem, he 
can always fall back on the direct laborious method. 

Before turning to the special devices, let us see how the method 
of joints may be applied without undue complication to the 
frame shown in Fig. 55a. This frame has 14 joints, and hence a 
direct attack involves 28 simultaneous equations. 

The load at M is W. We find at once (by taking components 
and moments) that the reactions at H and N are both vertical and 
of magnitudes R H = ^W, R N = W> Let SAB, Sac be the 
stresses in the bars. From the equilibrium of the joint AT, we 
have 



-fiir = - 



0. 



110 



PLANE MECHANICS 



[SEC. 3.5 



Passing to G, we have 



SQM sin a = SGN 
SFG = SGM COS a 



-%W cot a, 



where a is the inclination of the oblique bars to the horizontal. 
Proceeding in this way, step by step, we can find all the stresses. 
Incidentally, we shall get a check on our work when we reach the 
last joint. It will be noted that, to start the method, we must 
begin with a joint where only two bars meet. 

Method of sections. 

When we require the stresses in only some of the bars, the 
method of joints may prove unnecessarily laborious. Let us 
recall the fact, emphasized in Sec. 2.3, that we may choose any 
part of the given system as the system to which the conditions of 




FIG. 55fc. Method of sections: the "system" is enclosed by the broken line. 



equilibrium are applied. Up till now we have been thinking of 
a single bar, the whole frame, or a single joint as the system. 
But here we take a different approach, following the method of 
sections. 

Figure 556 shows the same frame as that of Fig. 55a, with the 
same load. We wish to find the stresses in KL, KE, DE. We 
consider as a system the part of the frame enclosed within a 
curved line cutting the bars KL, KE, DE, but no others. This 
system is acted on by the following external forces: 

the load W at M; 

the reaction RN at N; 

the stresses in KL, KE, DE. 

Taking moments about E, we have 

SKI EL + W - EF = RK ' EG. 



SBC. 3.5J APPLICATIONS IN PLANE STATICS 111 

But RN may be found by consideration of the equilibrium of the 
whole frame; hence 

SKL = \W COt a. 

From consideration of the total vertical component of force, we 
have 

SKE = ?W cosec a; 

and, from the total horizontal component, 

SDE SKL SKE COS a. = %-W COt a. 

We note that the method would not have worked had the three 
bars cut by the section met in a point. 

Method of virtual work. 

The method of virtual work may be applied to the problem just 
treated. To find SKL, we suppose the bar KL removed and 
forces applied to the joints K and L equal to the (unknown) stress 
in KL. The frame is no longer rigid, but it is in equilibrium. 
Hence the virtual work done in an infinitesimal displacement is 
zero. For infinitesimal displacement, let us take a rotation 
about E of the right-hand portion of the frame. The only forces 
to do work are the load W, the reaction RN, and the force at L 
replacing the stress SKL. Equating the work done by them to 
zero, we obtain the expression for SKL given above. 

To get the stress in EK, we replace the bar KL and remove 
EK, at the same time applying to the joints E and K forces equal 
to the (unknown) stress in EK. Now we give a virtual displace- 
ment, holding the left-hand portion fixed. The right-hand side 
rises slightly with parallel displacement of its bars, the bar DE 
hinging at D and KL hinging at K. The only working forces 
are the load W, the reaction R N , and the force at E replacing the 
stress SKE. Equating to zero the work done, we find for S K s 
the value given above. SDS is found similarly without difficulty. 

Complex frames. 

Frames constructed by adding successive pairs of bars to a 
basic triangular frame are called simple frames. Those so far 
discussed have been of this type. But there are also just- 
rigid frames which cannot be built up in this way; such frames 
are called complex. An example is shown in Fig. 56, in which 



112 



PLANE MECHANICS 



[SBC. 3.5 



the bars are supposed to cross without touching. The stresses 
may be found by solving the 12 equations of equilibrium of 

the joints, but that method is com- 
plicated. We cannot use the step- 
by-step method of joints, because 
there is no joint at which only two 
bars meet. We modify the method 
by assigning an unknown value to 
one of the stresses; we find the other 
stresses in terms of this one un- 
known by the conditions of equilib- 
rium of the joints and finally, on 
closing the calculation, determine 
the unknown stress and hence all the 
stresses. 

Let us work this out in the case 
shown in Fig. 56, in which the bars 
FE, ED, AD, FC are inclined to the 
horizontal at 45, and A B, BC in- 
clined to the horizontal at 30. Write SEB = S. Then 




FIG. 66. A complex frame. 



at E } 8*0 = Sn = 

at D, SAD = S*D = $/\/2, (= SFC, by symmetry), 

at Z>, SCD = (S, D - &UOA/2 = -S, 

at C, S CD + SFC/VZ + &BC/2 + Re = 0, 

at C, &c/\/2 + /SW-v/3/2 = 0. 

Elimination of SBC from the last two equations gives 
ScoV* + &c(V3 - 1)/V2 + RcV3 = 0. 
Since R c = W/2, SCD = -S, S,c = S/A/2, we obtain 

S =iTF(3 - V3); 
all the stresses are now easily written down. 

Concluding remarks. 

The methods described above are adequate in simple cases, 
and the reduction of the problem of determining the stresses 
to the solution of a set of 2j simultaneous linear equations is 
complete and satisfactory mathematically, although often com- 
plicated. When we have written down the equations, we know 
that we have given complete mathematical expression to all the 




SEC. 3.6] APPLICATIONS IN PLANE STATICS 113 

conditions of equilibrium and that the stresses can be found from 
the equations provided they are consistent. 

It may happen that the equations of equilibrium are incon- 
sistent; this occurs in the case of critical forms, of which an 
example is shown in Fig. 57. This 
frame is just-rigid ; but since the bars 
AB, EC lie in a straight line, no 
stresses in them can give equilibri- 
um of the joint B, when a load W is 
applied there. Such a frame would 
be an unsound engineering struc- 

turc. Actually the jointfl would be FlQ . 57 ._ A cnticid form . 
slightly depressed (owing to stretch- 

ing of the bars), and there would be very great tensions in the bars 
AB, BC. 

On account of its importance in engineering, the theory of 
frames has been elaborately developed. For a more complete 
account, with special reference to engineering problems, the 
reader is referred to S. Timoshenko and D. H. Young, Engineer- 
ing Mechanics (McGraw-Hill Book Company, Inc., New York, 
1940). 

Most statical problems admit two methods of attack. On the 
one hand, we may reduce the problem to the solution of equa- 
tions; this is the analytical method. On the other hand, we may 
represent forces by segments, and compound and resolve them 
by actual drawing; this is the graphical method. Each method 
has its advantages, but throughout this book we have preferred 
to use the analytical method, because it is easier to explain and 
has a wider range of application. For the graphical method in 
statics and its application to frames, the reader may consult 
for example H. Lamb, Statics (Cambridge University Press, 
1928). 

3.6. SUMMARY OF APPLICATIONS IN PLANE STATICS 
I. Mass centers and centers of gravity. 

(a) Formulas for the mass center: 



(3.601) r = ~ - (system of particles); 



114 PLANE MECHANICS [Sue. 3.6 

(3.602) f - V p f*f y f 8 (continuous system). 

JJjpdxdydz 

(6) Devices for finding mass centers: 

(i) symmetry, (ii) decomposition, (iii) theorems of Pappus. 

(c) Center of gravity coincides with mass center. Potential 
energy = Mgy. With respect to any vertical plane, the weights 
of all the particles of a system are plane-equipollent to a single 
force (total weight) acting through the center of gravity. 

II. Friction. 

Static friction: F/N ^ /* or 6 < X; (tan X = ju). 
Kinetic friction: F/N = n' or = X'; (tan X' = /*') 

III. Thin beams. 

(a) S 



-^T 



T = tension, S = shearing force, M bending moment. 
(6) Basic assumptions: 

(i) Hooke's law: T = k'e, (e = extension, &' = EA). 
(ii) Euler-Bernoulli law: M = k/p, 

(p = radius of curvature, k = El). 

(c) Differential equation of a thin heavy beam: 

(3.603) kj = -w. 

(3.604) If-jg, S=-f- 

(d) Continuity conditions: T/, dy/dx, d 2 y/dx* are continuous. 

IV. Flexible cables, 
(a) General formulas: 

(3.605) T d f a -H (a constant); (g) . J. 
(6) Cable of suspension bridge hangs in a parabola. 



Ex. Ill] APPLICATIONS IN PLANE STATICS 115 

(c) Common catenary: 

(3.606) y - c (cosh 5 - A c ^, 

(3.607) s = c sinh -> 

(3.608) s 2 + c 2 = (*/ + c) 2 , 

(3.609) r = 77 + wy. 

(d) Light cable in contact with a smooth curve: 

(3.610) T = constant, N = - 

P 

(e) Light cable in contact with a rough curve: 

(3.611) T = T &' (for cable on point of slipping). 
V. Frames. 

(a) Just-rigid frame: 

(3.612) b = 2 j - 3 

(b) Method of joints. Begin with a joint where only two bars 
meet. 

(c) Method of sections. Section must not cut more than three 
bars, and these three bars must not meet at a point. 

(d) Method of virtual work. Remove a bar. 

(e) Complex frames: (6) and (c) not applicable directly. 
Assume one stress <S, and use (6). 

EXERCISES III 

1. Find the mass center of a cubical box with no lid, the sides and bottom 
being made of the same thin material. 

2. A ladder leans against a smooth wall, the lower end resting on a rough 
floor for which the coefficient of friction is \. Find the inclination of the 
ladder to the vertical, if it is just on the point of slipping. 

3. A square frame is braced by two diagonal bars. One of these con- 
tains a turnbuckle, which is tightened until there is a tension T in the bar. 
Find the stresses in the other bars. 

4. A man of weight W walks slowly along a light plank of length a, 
supported at its ends. Find the bending moment in the plank directly 
beneath his feet as a function of his distance from one end of the plank. 
Find also the shearing forces j ust in front of him and just behind him. Draw 
diagrams to show the senses of the bending moment and shearing forces. 

5. Find the mass center of a wire bent into the form of an isosceles right- 
angled triangle. 

6. A rod 4 ft. long rests on a rough floor against the smooth edge of a 
table of height 3 ft. If the rod is on the point of slipping when inclined at an 
angle of 60 to the horizontal, find the coefficient of friction. 



116 PLANE MECHANICS [Ex. Ill 

7. A body of weight w rests on a rough inclined plane of inclination i, 
the coefficient of friction (/*) being greater than tan *. Find the work done 
in slowly dragging the body a distance a up the plane and then dragging it 
back to the starting point, the applied force being in each case parallel to 
the plane. 

8. A heavy cable rests in contact with a smooth curve in a vertical plane. 
Show that the difference in the tension at two points of the cable is propor- 
tional to the difference in level at these points. 

9. Two light rings can slide on a rough horizontal rod. The rings are 
connected by a light inextensible string of length a, to the mid-point of 
which is attached a weight W. Show that the greatest distance between 
the rings, consistent with the equilibrium of the system, is 



+ M 2 , 

where p is the coefficient of friction between either ring and the rod. 

10. A heavy beam ABCD, of weight 2 Ib. per ft., is supported horizontally 
by knife-edges at B and D. The beam is subjected to an additional vertical 
load of 20 Ib. at C. If AB = BC = CD = 4 ft., determine the shearing 
force and bending moment for all points of the beam. 

11. A portion of a circular disk of radius r is cut off by a straight cut of 
length 2c. Find the position of the mass center of the larger portion. 

If r 1 ft., c = 6 in., calculate the distance of the mass center from tho 
center of the circle. 

12. A light cable connects two weights W, w (W > w) and passes over a 
rough circular cylinder whose axis is horizontal. W rests on the ground, 
and w is suspended in the air. Find the least value of the coefficient of 
friction between the cylinder and the cable in order that W may be raised 
from the ground by slowly rotating the cylinder, and find expressions for 
the work done in turning the cylinder through one revolution (i) if W is 
raised, (ii) if W is not raised. 

Find also the work done in turning the cylinder through one revolution 
in the opposite sense. 

13. For the frame shown in Fig. 55a, take a = 45, and find the stresses 
in all the bars. Make a sketch of the frame, marking with a double line 
each bar in which there is a thrust. 

14. A uniform semicircular wire hangs on a rough peg, the line joining its 
extremities making an angle of 45 with the horizontal. If it is just on the 
point of slipping, find the coefficient of friction between the wire and the peg. 

15. An elastic beam rests on three props, two being situated at the ends 
of the beam and at the same height, and the third at the middle point of tho 
beam. Find the height of this central prop if the pressures on all three 
props are equal. 

16. A cable 200 ft. long hangs between two points at the same height. 
The sag is 20 ft., and the tension at either point of suspension is 120 Ib. wt. 
Find the total weight of the cable. 

17. A uniform cable hangs across two smooth pegs at the same height, 
the ends hanging down vertically. If the free ends are each 12 ft. long 



Ex. Ill] APPLICATIONS IN PLANE STATICS 117 

and the tangent to the catenary at each peg makes an angle of 60 with the 
horizontal, find the total length of the cable. 

18. Consider a frame as in Fig. 50, the triangles being equilateral. It is to 
carry a load 2W, either as a single load at E or equally divided between E 
and F. In which case is there greater danger of collapse, it being assumed 
that collapse is due to a thrust in a bar exceeding some definite value, the 
same for all bars? 

19. Four rods each of length a and weight w are smoothly jointed together 
to form a rhombus ABCD, which is kept in shape by a light rod BD. The 
angle BAD is 60, and the rhombus is suspended in a vertical plane from A. 
Find the tension or thrust in BD and the magnitude and direction of the 
force exerted by the joint C on the rod CD. 

20. Two equal spheres, each of weight W, rest on a horizontal plane in 
contact with one another. All three contacts are equally rough, with 
coefficient of friction p. The spheres are pressed together by forces of 
magnitudes P, Q (P > Q) acting inward along the line of centers. Show thn t 
there will be equilibrium if, and only if, 

P - Q <, n(P + Q), P - Q < A1W - (P - Q)]. 

If P and Q are increased, their ratio remaining fixed, how will equilibrium 
be broken? 

21. Find the stresses in the frame shown in Fig. 56 if the joint F is fixed, 
instead of A. 

22. A hanging cable consists of two portions for which the weights per unit 
length are w\ and Wz. Show that there is a discontinuity of curvature where 
the two portions are connected, the radii of curvature (pi, p 2 ) on the two sides 
of the join satisfying the equation p\w\ = pzWz. 

23. A beam AB, of length I and weight W, rests in a horizontal position 
with A clamped and a load W is suspended from B. If the weight per unit 
length of the beam varies as the square of the distance from B, show that at 
distance x from A the shearing force S and the bending moment M are given 
by 



W 
M - W'(l - x) + ~ (I - *)*. 

24. Prove that at a point inside a uniform solid sphere the force of attrac- 
tion varies directly as the distance from the center. 

25. Find the potential of a circular disk at a point on its axis. Use the 
result to calculate the potential of a solid sphere at an external point. 



CHAPTER IV 
PLANE KINEMATICS 

4.1. KINEMATICS OF A PARTICLE 

Having completed our study of plane statics, we now prepare 
for the study of dynamics by developing some results in kine- 
matics; this subject deals with the motions of particles and rigid 
bodies without any consideration of the forces required to produce 
these motions. In the present chapter we discuss kinematics 
in a plane. 

Tangential and normal components of velocity and acceleration. 
Consider a particle P moving in a plane, in which Oxy are 

fixed axes. The position vector of the particle (cf. Sec. 1.3) 


is r = OP, and the velocity is q = dr/dt. If the path of the 

particle is the curve (7, then dr is an 
infinitesimal displacement along C, so 
that 

dr = i ds, 

where ds is an clement of length on 
C and i is the unit vector tangent to 
C. Thus 




O * or, in words, the velocity of a moving 



FIG. 58a.--Resoiution along part id e h as a direction tangent to the 

tangent and normal. r 

path and a magnitude ds/dt. 

Let j be the unit normal vector to C (Fig. 58a), and let < be 
the inclination of i to the oxixis. As we move along C, i and j 
are functions of 0. Figure 586 shows the vectors i and i + Ai 
(corresponding to < and + A<, respectively) transferred to a 
common origin. Since these are both unit vectors, the triangle 
formed by the three vectors i, i + Ai, Ai is isosceles. The magni- 

118 



SBC. 4.1] PLANE KINEMATICS 119 

tude of Ai is 2 sin |A0, and so the limit of the magnitude of 
Ai/A0 is unity. Thus di/d<f> is a unit vector, pointing in the 
limiting direction defined by Ai as A< tends to zero. This direc- 
tion is clearly that of j, and so di/d<f> = j. When a similar 
argument is used to evaluate d]/d<f>, we readily see that dj/d<t> is 
perpendicular to j, but since the limiting direction of Aj is that 




FIG. 586. Change in unit tangent vector. 

of i, we get dj/d(j> = i. Combining these results we have 

/A ir\n\ dl . rfj 

(4.102) ^ = j, -I = -L 

To find the acceleration f, we differentiate (4.101); this gives 

(4103) | = *S. !*! + .*. * 

(1.W6) I df I ^ -r- ^ d 

Hence, by (4.102) and the fact that the radius of curvature of 
C is p = ds/d<t>, we have 

(4.104) f = i S + i ^ ; 

or, in words, the acceleration of a moving particle has a component 
dq/dt along the tangent and a component q*/p along the normal to 
the path. The tangential component may also be expressed in 
the form q dq/ds. 

It is easily seen that the normal component of acceleration 
always points to the concave side of the path. 

As an example, consider a particle traveling in a circle of radius r with a 
speed q which is (a) constant, (6) proportional to t. In case (a), the accelera- 
tion vector is directed inward along the radius and has a magnitude g a /r; 
in case (6), the acceleration vector has a constant component along the 
tangent and a component along the radius which varies as '. 



120 



PLANE MECHANICS 



[SEC. 4.1 



Radial and transverse components. 

Consider a particle P moving in a plane, its position being 
described by polar coordinates r, 6 (Fig. 59). Let i be the unit 
vector along OP and j the unit vector perpendicular to i, drawn 

in the sense shown. We note that 




x 

FIG. 59. Resolution along 
and perpendicular to radius 
vector. 



(4.107) 
Thus 



(4-105) 



de 



We have then r = ri, and the velocity 
is 

(4.106) q = f = ri + rtj, 

the dot indicating d/dt. Thus (f, r6) 
are the components of velocity along 
and perpendicular to the radius vector. 
For the acceleration we find, on 
differentiating (4.106) and using 
(4.105), 



are the components of acceleration along and perpendicular to the 
radius vector. 

It is usual to call the components in the directions i and j 
the radial and transverse components, respectively. 

The hodograph. 

It is easy to form an intuitive picture of the velocity of a 
particle; we have merely to visualize the small displacement 
it receives in a small time and then imagine that small displace- 
ment greatly magnified without change of direction. But it is 
much more difficult to form an intuitive picture of the accelera- 
tion. The hodograph is a device to facilitate this. Figure 60a 
shows the path C of a particle, with its velocity q and acceleration 
f at the position P. Imagine now a fictitious particle P' moving 
in the plane (Fig. 606) with a motion correlated to the motion 



SEC. 4.2] 



PLANE KINEMATICS 



121 



of P by the following rule: the position vector of P', relative to 
some chosen origin 0', is equal to the velocity of P. The path 
described by P' is called the hodograph of the motion of P. 





0' 

(a) (6) 

FIG. 00. (a) Motion, (b) Hodograph. 

Denoting by r', q' the position vector and velocity of P', we 
have 

(4.108) r' = q. 
Hence, on differentiation, 

(4.109) q' = f; 

in words, the velocity in the hodograph in equal to the acceleration 
in the actual motion. 

Exercise. Verify the following statements: 

(i) If a particle has an acceleration which is constant in magnitude and 
direction, the hodograph is a straight line described with constant speed. 

(ii) If a particle moves in a circle with constant speed, the hodograph is a 
circle described with constant speed. 

4.2. MOTION OF A RIGID BODY PARALLEL TO A FIXED PLANE 
Description of the motion. 

As already remarked in Sec. 2.4, the motion of a rigid body 
parallel to a fixed plane is completely described by the motion 
of the representative lamina, i.e., the section of the body by 
the plane. We may therefore confine our attention to the 
representative lamina. We also discussed in Sec. 2.4 the general 
infinitesimal displacement of a lamina in its plane. This dis- 
placement was described by selecting a base point A in the lamina 




122 PLANE MECHANICS [Sac. 4.2 

and giving (i) the infinitesijnal displacement of A and (ii) the 
infinitesimal angle through which the lamina is turned. 

A continuous motion of a lamina may be considered as a 
sequence of infinitesimal displacements received in infinitesimal 
intervals of time. We select some particle A of the lamina as 
base point. At any time t, A has a velocity, say q^. At time t 
the angle between a line fixed in the lamina and a line fixed in the 
plane is increasing at some rate which we shall denote by : 

w is called the angular velocity of the 
lamina.* In a small time interval 
dt the particle A receives a small 
displacement q^ dt, and in the same 
interval the lamina is turned through 
a small angle w dt. Hence the 
specification of q A and co as functions 
of t describes the succession of 
infinitesimal displacements which 
the body undergoes. To sum up: 
x The motion of a lamina in a plane is 

Fio.61. The motion of a lamina 7 ., , 7 ,.^ , .. * 

described by <u and *>. described by (i) selecting a base point 

A in the lamina , (ii) specifying the 

velocity q^ of A as a function of the time t, and (iii) specifying the 
angular velocity co of the lamina as a function of t. 

This description, for the instant t, is shown diagrammatically in 
Fig. 61, the curved arrow being used to indicate angular velocity. 

The instantaneous velocity of any point P of the lamina can 
be found from (2.409), which gives the infinitesimal displacement 
of a point. Let (a, 6) be the coordinates of A and (x, y) those 
of P, both measured on fixed axes Oxy. Then the infinitesimal 
displacement of P in the time interval dt has components 

dx = u * di ~ (y ~ 6 ) w dt > 
- a)co dt, 

where U A , V A are the components of q^. Thus the velocity of P 
has components 

u - tu - fc, - 



(4.202) 

( v = VA + (x a)co. 

* Since the angle between two lines fixed in a rigid body is constant, it is 
easily seen that the value of w is the same no matter what lines are chosen in 
the lamina and in the plane. 



SBC. 4.2) PLANE KINEMATICS 123 

Instantaneous center. 

At any instant, there is just one point of a moving lamina which 
has no velocity. Its coordinates are found from (4.202), on 
putting u = v = 0; they are 



(4.203) 



x = a 



y = b + UA/W- 



This point is called the instantaneous center. Under one excep- 
tional condition no instantaneous center exists, namely, when 
co = 0. We may then say that the instantaneous center is at 
infinity. 

C 





FIG. 62. Determination of the 
instantaneous center from the 
velocities of two points. 



FIG. 63. The body centrode B 
rolls on the space centrode S. 



Once the instantaneous center C is known, it is very easy to 
visualize what happens to the lamina in a small interval of time 
dt: the lamina rotates about C through a small angle co dt. This 
fact enables us to find C when the directions of the velocities of 
two points A and B of the lamina are known. For, since the 
lamina is turning about C at the instant, the velocity of any 
point P is perpendicular to CP. Hence, C is located at the 
intersection of the lines drawn through A and B perpendicular 
to the velocities of those points (Fig. 62). 

From the definition of the rolling of one curve on another, 
given in Sec. 2.4, it is now clear that a moving curve rolls on a 
fixed curve when the curves touch and the instantaneous center of 
the moving curve is at the point of contact. In fact, this statement 
may be taken as a definition of rolling, instead of that given in 
Sec. 2.4. If both curves are in motion, we define rolling by the 
conditions that the curves touch and that the instantaneous 



124 PLANE MECHANICS [SBC. 4.2 

velocities of the two particles at the point of contact (one on 
each curve) are equal to one another. 

As a lamina moves, the instantaneous center C moves in the 
fixed plane; the curve described by it is called the space centrode 
(S). But C also moves in the lamina; the curve described by C 
in the lamina is called the body centrode (B). At any instant, 8 
and B have the point C in common, and B is turning about (7, 
since B is carried along with the lamina. The situation a time t 
is shown in Fig. 63. A little later, at time t + dt, a point D of B 
has moved into coincidence with a point E of S to form the new 
instantaneous center, and this is done by turning B about C 
through the small angle w dt. Hence, it is evident that B cannot 
cut S at a finite angle; therefore B touches S at (7, and since C is 
the instantaneous center of B, we have the following result: The 
body centrode rolls on the space centrode. 

Exercise. Verify the following statements: 

(i) When a wheel rolls on a track, the space centrode is the track itself 
and the body centrode the circumference of the wheel. 

(ii) When a rod of length 2a slides with its extremities on two lines which 
intersect at right angles, the body centrode is a circle of radius a and the 
space centrode a circle of radius 2a. 

Example. As an example of our analysis of the motion of a rigid body, 
let us consider two wheels Wi and Wz of radii ai and az, respectively, lying 
in a plane. Their centers are connected by a rod R of length ai + 2, and 
the wheels engage without slipping. If we fix the center of Wi, then W\ and 
R can turn independently about this center, and Wz will roll on W\. Each of 
the three bodies W\, Wz, R has an angular velocity, say coi, 2, ft. These 
three angular velocities are not independent; let us find the relation con- 
necting them. 

The particles of W\ and Wz at their point of contact have the same veloc- 
ity. We can find two different expressions for this common velocity; equat- 
ing them, we obtain the required relation. Since Wi turns about its center 
with angular velocity coi, the velocity of its particle at the point of contact is 
tangential and of magnitude anu\. The wheel Wz has a motion which may 
be described by means of a base point taken at its center. The velocity 
of this base point is perpendicular to R and of magnitude (cti -f aa)S2. 
Hence the velocity of the particle of W* at the point of contact with Wi is 
tangential and of magnitude 

(ai + az) ft a 2 co 2 . 
Therefore we have, as the required relation in symmetric form, 



Ex. IV] 



PLANE KINEMATICS 



125 



As an alternative method of finding w 2 when wi and ft are given, the follow- 
ing general method of finding angular velocity may be used. Take two 
particles of the body, say A and B. Resolve their velocities perpendicular 
to AB. The difference of these component velocities, divided by AB, is the 
required angular velocity. The proof of this is left as an exercise. 

4.3. SUMMARY OF PLANE KINEMATICS 
I. Kinematics of a particle. 

(a) Components of velocity and acceleration: 





Velocity 


Acceleration 


Tangential . 
Normal 


q = s 



dq 
SOrq Ts 

t 


Radial 


T 


P 
r - r6* 


Transverse .... 


r6 


i>-> 



(6) Position in hodograph is velocity in motion; velocity in 
hodograph is acceleration in motion. 

II. Kinematics of a rigid body. 

(a) The angular velocity co of a lamina is the rate of change 
of the angle between a line fixed in the lamina and a line fixed 
in the plane of reference. 

(6) For base point A (a, 6) the velocity at (x, y) has components 

(4.301) u = U A - (y - 6), v = VA + (x - a)co. 

(c) The space centrode (S) is the locus of the instantaneous 
center in the plane of reference. The body centrode (B) is the 
locus of the instantaneous center in the body. B rolls on S. 
EXERCISES IV 

1. A particle moves in a plane with constant speed. Prove that its 
acceleration is perpendicular to its velocity. 

2. A particle moves in an elliptical path with constant speed. At what 
points is the magnitude of the acceleration (i) a maximum, (ii) a minimum? 

3. A particle moves along a curve y = a sin px, where a and p are con- 
stants. The component of velocity in the ^-direction is a constant (w). 
Find the acceleration, and describe the hodograph. 

4. AB, BC are two rods, each 2 ft. long, hinged at B. A and C are made 
to slide in a straight groove in opposite directions, each with a speed of 8 ft. 



126 PLANE MECHANICS [Ex. IV 

per sec. Find the velocity and acceleration of B at the instant when the 
rods are perpendicular to one another. 
6. Starting from 

x * r cos 0, y r sin 0, 

calculate $ and #. Hence, by resolving the acceleration vector along and 
perpendicular to the radius vector, establish the formula (4.107). 

6. A wheel of radius a rolls without slipping along a straight road. If 
the center of the wheel has a uniform velocity v, find at any instaiit the 
velocity and acceleration of the two points of the rim which are at a height h 
above the road. Examine in particular the cases h = 0, h = 2o. 

7. Is it possible for a particle to move in a circle and have a hodograph 
which is a straight line? Give reasons for your answer. 

8. A wheel of radius a rolls along a straight track, the center having a 
constant acceleration /. Show that there is, at any instant, just one point 
of the wheel with no acceleration; find its position relative to the center of 
the wheel. 

9. A rectangular plate A BCD moves in its plane with constant angular 
velocity . At a given instant the point A has a velocity of magnitude V 
along the diagonal AC. Find the velocity of B at this instant in terms of 
V, u>, and the dimensions of the rectangle. 

10. A uniform circular hoop of radius a rolls on the outer rim of a fixed 
wheel of radius 6, the hoop and the wheel being coplanar. If the angular 
velocity u of the hoop is constant, find 

(i) the velocity and acceleration of the center of the hoop; 
(ii) the acceleration of that point of the hoop which is at the greatest 
distance from the center of the wheel. 

11. A motorboat experiences a resistance proportional to the square of the 
speed. The engine is switched off when the speed is 50 ft. per sec. When 
the boat has moved through a distance of 60 ft., its speed has been reduced 
to 20 ft. per sec. Find (to the nearest foot) the total distance traversed 
when the speed has been reduced to 10 ft. per sec. 

12. A point A has a uniform circular motion about a fixed point with 
angular velocity m. A point B has a uniform circular motion about A with 
angular velocity n. What relation connects m and n if the acceleration of 
B is always directed toward 01 

13. A circular ring of radius 6 turns in its plane about its center with 
constant angular velocity ft. A second circular ring of radius a (<&) rolls 
in the same plane on the inner side of the first ring. The angular velocity 
of the center of the smaller ring about the center of the larger ring is , a 
constant with the same sign as 0. Find the space and body centrodes for 
the smaller ring. 

14. A particle P moving in a plane has an acceleration directed toward a 
fixed point in the plane and varying as I/OP 2 . Show that the curvature 
of the hodograph is constant and hence that the hodograph is a circle. 



CHAPTER V 



METHODS OF PLANE DYNAMICS 

6.1. MOTION OF A PARTICLE 
Equations of motion. 

In accordance with (1.402) a particle, under the influence 
of a force P, moves so as to satisfy the equation 

(5.101) mi = P, 

where m is the mass of the particle and f its acceleration rela- 
tive to a Newtonian frame of reference. If Oxyz are rectangu- 
lar axes in this frame, then (5.101) gives, on resolution into 
components, 

(5.102) mx = X, my = Y, mz = Z, 

where X, Y, Z are the components of P along the axes. 

The above statements hold for a par- 
ticle moving in space; let us now confine 
our attention to a particle moving in a 
plane, the force P being supposed to act 
in the plane of motion. 

We may resolve P into components 
along the tangent and normal to the path 
of the particle (Fig. 64). If these com- 
ponents are P, P n , respectively, the 
vector equation of motion (5.101) gives, 
on resolution along the tangent and 
normal Icf. (4.104)], 

(5.103) "S?- P * ?-'-*- 

An interesting deduction may be noted. If the force acting 
on a particle is always perpendicular to its velocity (so that 
P t = 0), then the speed of the particle is constant. If, further, 
the force is of constant magnitude, then P n is constant; then 

127 




O x 

Fio. 64. Resolution of 
force along tho tangent 
and normal to the path. 



128 PLANE MECHANICS [SBC. 5.1 

p is constant, so that the particle describes a circle. This occurs 
when an electrically charged particle moves in a uniform mag- 
netic field with lines of force perpendic- 
ular to the plane of motion. 

Let us now consider a particle mov- 
ing in a plane under the influence of a 
force always directed away from, or 
toward, the origin (Fig. 65). Such a 
force is called a central force. Let R 




O x be the component of force in the direc- 

Fio. 05.-A central force. away from ^ ^^ ^ ^^ R 



positive when the force is repulsive and negative when it is 
attractive. Then, by (5.101), on resolution along and perpen- 
dicular to the radius vector [cf. (4.107)], 

(5.104) m(r - r0 2 ) = R, ^ (r 2 d) = 0. 

Equations (5.102), (5.103), and (5.104) are all very useful 
forms of the equations of motion of a particle. 

Exercise. Referring to (3.122), write down the equations of motion of 
a particle in the earth's gravitational field, using (i) polar coordinates and 
(ii) rectangular Cartesians. 

Principle of angular momentum. 

The momentum of a particle of mass w, moving with velocity 
q, is defined as the vector mq. (This is sometimes called linear 
momentum, to distinguish it from angular momentum, defined 
below.) The components of momentum for motion in a plane 
are 

mx, my, 

where Oxy are rectangular axes in the plane. Since momentum 
is a vector, it has a moment about any point A in the plane; this 
moment is called moment of momentum or angular momentum 
about A. In Chap. II, we discussed moments of vectors; we 
saw that the moment of a vector is the sum of the moments of 
its components, and so by (2.303) the angular momentum of a 
moving particle about the origin is 

(5.105) h = m(xy yx). 

If, instead of resolving the momentum vector along the axes, we 



SEC. 5.1] METHODS OF PLAXE DYNAMICS 129 

resolve it along and perpendicular to the radius vector drawn 
from the origin, we obtain components [cf. (4.10(5)] 

mr, mrd. 

The former component has no moment about the origin; hence 
(5.106) h 



Consider now a particle moving in a plane under the action 
of a force with components X, Y. The rate of change of angular 
momentum about the origin is 

h = m(xy yx) = xY yX = JV, 

where N is the moment of the force about the origin. Hence we 
have the principle of angular momentum: For a particle moving in 
a plane the rate of change of angular momentum about any fixed 
point in the plane is equal to the moment of the force about that 
point. 

We note that, in the case of a central force, the second equation 
of (5.104) is equivalent to the statement that the angular momen- 
tum about the origin is constant. 

Linear momentum, being the product of mass and velocity, has the dimen- 
sions [MLT~ 1 ]. In the c g s. system, it is measured in gin. cm. sec." 1 ; in the 
f p.s. system in Ib. ft. sec." 1 On account of the equivalence of dimensions 
(see Appendix), these units may also be called dyne sec. and poundal sec., 
respectively. Angular momentum has the dimensions [ML Z T~ 1 ] and is 
measured in gm. cm. 2 sec.~ l or Ib. ft. 2 sec." 1 

Principle of energy. 

The kinetic energy* of a particle of mass m moving with velocity 
q is defined to be %mq 2 . It will be denoted by T. Thus, for a 
particle moving in space, 

(5.107) T = frnq 2 = ?m(x 2 + y* + 3 2 ). 

The rate of change of kinetic energy is 

T = m(xx + yy + zz) = Xx -f- Yy + Zz, 

where X, Y, Z are the components of the force acting on the 
particle. The increase in kinetic energy in the interval (t , ti) 

* Dimensions (ML*T~*], as for work or potential energy and measured in 
the same units (cf. p. 54). 



130 PLANE MECHANICS [SBC. 5.1 

is therefore 

r x - To - (Xx + Yy + Zz) dt. 

Let W denote the work done by the force during this time 
interval. By (2.403) the work done in an infinitesimal displace- 
ment is 

dW X dx + Y dy + Z dz = (Xx + Yy + Zz) dt, 
and so 

W = J[" (Xx + Yy + Zz) dt. 
Hence 

(5.108) Ti - To = W. 

This establishes the principle of energy: The increase in kinetic 
energy is equal to the work done by the force. 

Differentiating (5.108) with respect to t\ and then dropping the 
subscript 1, we have 

(5.109) T = W; 

in words, the rate of increase of kinetic energy equals the rate of 
working of the force. 

If the particle moves in a conservative field of force with 
potential energy V, then, as in (2.419), 

v dV v dV dV 

Zs3 -ar F ~~V z =-Tz' 

Then 

W = f 1 (Xx + Yy + Zz) dt = - V dt = -Ft + 7 , 

where V\ y VQ are the potential energies at times t\, t , respectively. 
Comparison with (5.108) gives 

T, - To - -.7i> 7o, 2 1 ! + 7i - To + V . 
Hence, in general, 

(5.110) T + V - E, 

where E is a constant, called the total energy. Thus the sum of 
the kinetic and potential energies is constant. This is called the 
principle of the conservation of energy. 
The principle expressed mathematically by (5.110) is one of 



SEC. 5.2] METHODS OF PLANE DYNAMICS 131 

the fundamental formulas of mechanics, and it is of great use in 
the solution of problems. It represents one relation among the 
three coordinates and the three components of velocity, V being 
supposedly known as a function of the coordinates. In the 
absence of a conservative field, we no longer have (5.110), only 
(5.108). This is much less useful because W is not a function of 
the coordinates. It is an integral the value of which depends 
on the path of the particle and thus is unknown, since the path of 
the particle is precisely what we have to find in the majority of 
problems on the dynamics of a particle. 

Exercise. A particle slides down a smooth inclined plane. Use (5.110) 
to find its speed in terms of the distance traveled from rest. 

6.2. MOTION OF A SYSTEM 

Anyone familiar with the usual type of problems posed as 
exercises in mechanics must have been struck by their artificial 
character. The mechanical systems considered are often too 
simple to be of much practical interest. Attention is concen- 
trated on such simple systems as rigid bodies swinging about 
fixed axes or wheels rolling along lines, instead of on complicated 
realities like trains, automobiles, or airplanes. This is because 
it has become traditional in the study of mechanics to direct 
attention to problems which are soluble, in the sense that the 
behavior of the system can be described by simple formulas. 
This is an unfortunate practice, because it fails to emphasize one 
of the greatest achievements of the applied mathematician, 
namely, his capacity to make general statements about com- 
plicated systems without paying much attention to the details 
of the systems. 

To extract the fullest interest from the present section, the 
reader should bear in mind the striking generality of the state- 
ments. Since, however, it is tiring and confusing to think too 
much in terms of generalities, he should bear in mind a few 
concrete examples and think of them in connection with the 
various principles about to be discussed. The following systems 
are suggested as suitable examples: 

(i) a stick sliding on a frozen pond; 

(ii) a complete automobile; 

(iii) a wheel of an automobile; 



132 PLANE MECHANICS [SEC. 5.2 

(iv) an airplane; 

(v) a man on a trapeze; 

(vi) the solar system. 

As attention is at present directed toward plane dynamics, it 
is advisable to think primarily of two-dimensional motions of 
the above systems; e.g., the automobile and the airplane are 
traveling straight ahead. For the first five systems, we may 
accept the earth's surface as a Newtonian frame. As for the 
solar system, we may merely assume that there is some New- 
tonian frame and try to identify it by examining the consequences 
of the laws of motion. 

The system under consideration is regarded as composed of 
particles. The forces acting on the particles are in part internal 
and in part external, the internal forces satisfying the law of 
action and reaction (Sec. 1.4). A rigid body is a particular type 
of system, in which the internal forces are such as to prevent the 
alteration of the distances between the particles. The internal 
forces in a system are, as a rule, complicated; the purpose of 
the general principles which we are about to establish is to make 
important statements about the motion of the system which 
involve, not these complicated forces, but only the external 
forces which as a rule are comparatively simple. For example, 
in the case of the automobile, the only external forces are (i) 
gravity, (ii) reactions at the contacts of the tires with the ground, 
and (iii) resistance of the air. 

Principle of linear momentum; motion of the mass center. 

The linear momentum of a system is defined as the sum of the 
linear momenta of the several particles of the system. Thus, 
if the masses of the particles are mi, w 2 , m, and* their 
velocities qi, q2, q, the linear momentum of the system is 
the vector 

n 

(5.201) M = 2) m '<l" 

t i 

We shall now prove the principle of linear momentum: The 
rate of change of linear momentum of a system is equal to the vector 
sum of the external forces. 

From (5.201), we have 

(5.202) M = mA, 



SEC. 5.2] METHODS OF PLANE DYNAMICS 133 

where f t is the acceleration of the ith particle. Thus, 

(5.203) & = (P, + PO, 

= 1 

where P is the external force on the ith particle and Pj the 
internal force on it. But, from the equality of action and reac- 
tion, we see that 

(5.204) p; = o, 

1 = 1 

because this summation consists of vectors which are formed 
from pairs of forces that are equal and opposite. Hence, (5.203) 
gives 

(5.205) M = J) P,, 

t= 1 

which proves the principle of linear momentum. 
The last equation may be written 

(5.206) M = F, 

where F is the vector sum of the external forces. 

We shall now prove the law of motion of the mass center: The 
mass center of a system moves like a particle, having a mass equal 
to the total mass of the system, acted on by a force equal to the vector 
sum of the external forces acting on the system. 

From (3.101) it follows that the velocity of the mass center 
of a system of particles is 

n 

(5.207) q - X MUM 

t=i 

where m is the total mass of the system. Thus if f is the accelera- 
tion of the mass center, we have 

(5.208) ml = w4 
and so, by (5.206), 

(5.209) mi = F. 

This is the equation of motion of a particle of mass m acted on by 
a force F, and so the law is established. 



134 PLANE MECHANICS [Sac. 5.2 

We note that, by (5.207), 

n 

(5.210) mq - ] m t q t -, 



so that the linear momentum of the fictitious particle moving 
with the mass center is equal to the linear momentum of the 
system. 

The conclusions to be drawn from the preceding principles 
are simple and interesting when the vector sum of the external 
forces is zero. Then the linear momentum of the system remains 
constant, and its mass center travels in a straight line with 
constant speed. This is true in particular for a stick sliding on a 
frozen pond. As for the solar system, we see that any New- 
tonian frame of reference must be such that the mass center of 
the solar system has a constant velocity relative to it. 

Principle of angular momentum; motion relative to mass center. 

The angular momentum of a system about a line (or about a 
point in its plane if the system is confined to a plane) is defined 
as the sum of the angular momenta of the particles composing it. 
Thus if the particle of mass m* has coordinates # t , y^ z % and 
velocity components &, #, z,, the angular momentum of the 
system about Oz is 



(5.211) h ] 

~i 

The rate of change of angular momentum about Oz is then 

n 

(5.212) 



=! 

But 

m&i X. + X'i, mjj v = Y t + Y' it 

where Xi, Yi are the components of external force acting on the 
particle and X f i} FJ the components of internal force. Thus, 



(5.213) h - 2 & Y < ~ y^ + 2 
- ffi ffi 

N + N', 

where -AT is the total moment about Oz of all the external forces 
acting on the system and N' the total moment of all the internal 



SBC. 5.21 METHODS OF PLANE DYNAMICS 135 

forces. But since the internal forces occur in balanced pairs, 
their total moment is zero; hence 

(5.214) h - N. 

This equation expresses the principle of angular momentum: 
The rate of change of the angular momentum of a system about 
a fixed line is equal to the total moment of the external forces about 
that line. 

If we think, for example, of a man on a trapeze, the only 
external forces are (i) gravity and (ii) a reaction at the point of 
suspension. But the latter has no moment about the point of 
suspension. Hence the rate of change of angular momentum 
about the point of suspension is equal to the moment of the 
gravitational forces about that point. 

Let x, y t z be the coordinates of the mass center of a system 
and let #(, y{, z( be the coordinates of the ith particle relative to 
the mass center, so that 

(5.215) x, = * + *J, y< = y + y(. 

The components of velocity of the particle relative to the mass 
center are x f it y(, z(, and so the angular momentum relative to a 
line through the mass center parallel to Oz is 

n 

(5.216) h = % m v (x(y( - y(x(\ 



where we use, in computing angular momentum, the velocities 
relative to the mass center. We shall refer to this briefly as the 
angular momentum relative to the mass center. 
From (5.216), we obtain 



(5.217) 

=i 

and hence by (5.215), differentiated twice, 

(5.218) h = 






136 PLANE MECHANICS [SEC. 52 

as before, X t , F t are components of external force and X{, Y( 
components of internal force. But, from the defining property 
of the mass center, we have 

m % x( - 

so that the first two terms on the right-hand side of our last 
equation vanish. The fourth term vanishes through the balanc- 
ing of the internal forces in pairs, and so We have 

(5.219) h = N, 

where N is the total moment of the external forces about the 
mass center. This is the principle of angular momentum relative 
to the mass center: The rate of change of angular momentum relative 
to the mass center is equal to the moment of the external forces about 
the mass center. 

Exercise. Check to see that the two sides of (5.219) have the same 
dimensions. 

It will be noticed that we have a principle of angular momen- 
tum relative to a fixed axis and a principle of angular momentum 
relative to the mass center. The principle does not hold for an 
arbitrarily moving axis with fi,xcd direction. 

As an illustration of the principle of angular momentum relative to the 
mass center, consider the front wheel of an automobile. The external forces 
on it are (i) gravity, (ii) the reaction of the axle, and (iii) the reaction of 
the ground. The force of gravity and the reaction of the axle have no 
moment about the central line of the axle, which passes through the mass 
center. Hence the rate of change of angular momentum relative to the 
center of the wheel equals the moment of the reaction of the ground about 
the center. In particular, if the car is traveling at constant speed, the 
angular momentum is constant, and so the reaction of the ground must 
act vertically up through the center of the wheel; no force of friction is called 
into play. Further, if the wheel bumps off the ground, its angular momen- 
tum will remain constant as long as it is in the air. These statements are 
made on the assumption that the bearings are smooth; the reader can 
supply the qualitative description of the modifications which arise when 
there is friction in the bearings. 

The principle of energy. 

The kinetic energy of a system is defined as the sum of the 
kinetic energies of its constituent particles; the formal expression 



SEC. 5.2] METHODS OF PLANE DYNAMICS 137 

is 

(5.220) T = 1 |) m t (x t 2 + + *J). 
Then, 

(5.221) T = V ^(iA + frfr + A) 

1=1 



where Xi, F-, Zi are the components of the total force, external 
and internal, acting on the ith particle. Thus, if W is the work 
done by the forces from time to to time t } we have 

(5.222) T = W. 

This is formally the same as (5.109), but here we are con- 
sidering a system instead of a single particle. For a system, 
the principle of energy takes the following form: The rate of 
change of kinetic energy of a system is equal to the rate of working 
of all the forceSj external and internal. 

There is a sharp difference between the principles of linear 
and angular momentum on the one hand and the principle of 
energy on the other. In the principles of momentum the internal 
forces are eliminated; in the principle of energy they are not 
eliminated, except in the special case where they do no work and 
so contribute nothing to W. In our idealized mathematical 
models, consisting of rigid bodies with smooth contacts, no work 
is done by the internal forces, and so they disappear from the 
principle of energy. In cases of collision, however, work may 
be done by the internal forces (see Chap. VIII) ; that is because, 
in such cases, it is impossible to regard the bodies as absolutely 
rigid. 

When the system is conservative, with potential energy V, 
we have W = - V by (2.416); then (5.222) leads to the principle 
of the conservation of energy 

(5.223) T + V = E, 
where E is the constant total energy. 



138 PLANE MECHANICS [Ssc. 5.2 

D'Alembert's principle. 

The principle about to be discussed adds nothing essential 
to the principles already given, but it is interesting as an alter- 
native expression. 

We have regarded "force " as a primitive concept in mechanics, 
and we shall not abandon that point of view. One must guard 
against logical confusion in accepting the following definitions, 
in which we use the conventional terms. Consider a particle of 
mass m t having at a certain instant an acceleration f. The 
vector mf is called the "effective force " acting on the particle, and 
that vector reversed, i.e., mf, the "reversed effective force." 
Now consider a system ($) of n particles in motion, the reversed 
effective force on the ith particle being mdi. Alongside the 
mental picture of this system, think of another ($') in which 
the particles are at rest at the same positions as they have 
instantaneously in S and are acted on by the same forces, external 
and internal, as in S; in addition let there act in the statical 
system S' a set of real forces identical with the reversed effective 
forces of S. Now, by the equations of motion of the particles 
in Sj we have 

Pi - m& = 0, (i = 1, 2, - - . n), 

where P is the real force on the ith particle in S', hence it follows 
that S' is in statical equilibrium since the total force on each 
particle is zero. Thus we have D'Alembert's principle: The 
reversed effective forces and the real forces together give statical 
equilibrium. 

To deal with problems in plane dynamics, we introduce a 
fundamental plane to which the motion is parallel. Since the 
internal forces are plane-equipollent to zero, it follows that 
the external forces, together with the reversed effective forces, form a 
system plane-equipollent to zero. We recall that this means that 
the vector sum and the moment vanish. 

The statement of D'Alembert's principle may give the impres- 
sion that it reduces dynamics to statics. This is partly true in 
the sense that the statement involves only the conditions of 
statical equilibrium. However, it must be remembered that the 
reversed effective forces involve derivatives of coordinates and 
that therefore conditions of statical equilibrium involving these 
forces are actually differential equations of motion. To determine 



SEC. 5.3] 



METHODS OF PLANE DYNAMICS 



139 



motion under given forces, these differential equations must be 
solved a dynamical, rather than a statical, problem. On the 
other hand, if the motion is known, D'Alembert's principle 
enables us to use the methods of statics to determine the forces 
acting on the system. 

Exercise. Three equal particles are joined by light rods to form an equi- 
lateral triangle. If the triangle rotates in its plane about its centroid with 
constant angular velocity, find the tensions in the rods. 

5.3, MOVING FRAMES OF REFERENCE 

In developing dynamics up to this point, we have assumed 
the existence of a frame of reference relative to which bodies 
move in accordance with the Newtonian laws. To get accurate 
agreement between theoretical prediction and observation, we 
take for frame of reference one in which the mass center of the 
solar system is fixed and which has no rotation relative to the stars 
as a whole. For a slightly less accurate agreement in the case 
of experiments on the earth, we may take the earth itself as 
frame of reference. 

We now raise the question: Knowing that a body behaves 
relative to a Newtonian frame of reference in accordance with 
the laws and principles discussed earlier, how does a body appear 
to behave when viewed from a frame 
of reference moving relative to the 
Newtonian frame? 



O 



y' 



s 



s' 



x f 



Frames of reference with uniform 
translational velocity. 

Let S be a Newtonian frame of 
reference and S' a frame of reference 
which has, relative to S, a uniform 
(i.e., unaccelerated) translational mo- 
tion. (If S is the earth's surface, ' 
might be a train running smoothly on 
straight tracks at constant speed.) We shall consider only two 
dimensions, but the argument can be extended to space 
immediately. 

In S we take axes Oxy and in S' we take parallel axes O'x'y' 
(Fig. 66). Let , rj be the coordinates of 0' relative to 0. Then 



FIG. 66. Frames of refer- 
ence in relative motion with- 
out rotation. 



(5.301) 



wo, 



t>o, 



140 PLANE MECHANICS [Sjuc. 5.3 

where u , v are the constant components of the velocity of S' 
relative to S. Let A be the position of any moving particle; it 
has coordinates (x, y) relative to Oxy and coordinates (#', y') 
relative to O'x'y'. These coordinates are connected by the 
relations 

x = x r + , y = y' + 97, 

and, on differentiation, 

(5.302) x - x' + , y = y r + 17. 

If we denote by q the velocity of A relative to S, by q' the velocity 
of A relative to S', and by q the velocity of S' relative to AS', 

(5.302) may be expressed in the form 

(5.303) q = q' + q<>. 




This is called the law of composition of 
velocities and is exhibited graphically in 

FIG. 67. Composition Fig. 67. 
of velocities. 

Exercise. A man stands on the deck of a 

steamer, traveling east at 15 miles per hour. To him the wind appears to 
blow from the south with a speed of 10 miles per hour. What is the true 
speed and direction of the wind? 

Since j, ij are constants, differentiation of (5.302) gives 

(5.304) x = *', y = y'; 

thus the acceleration relative to S' is equal to the acceleration 
relative to S, and this may be expressed in vector form as f ' = f . 
Thus the law of motion 

(5.305) mi = P 
may also be written 

(5.306) mi' = P, 

and so Newton 1 s law oflmotion holds in S' as well as in S. 

From this we draw an important conclusion. Given one 
Newtonian frame of reference S, we can find an infinity of other 
Newtonian frames of reference, namely, all those frames of 
reference which have a uniform motion of translation relative 

tO AS. 

Tn Sec. 5.2 we saw that if a Newtonian frame exists and of 



SEC. 5.3] METHODS OF PLANE DYNAMICS 141 

course we suppose that it does, since otherwise* there would be 
no Newtonian mechanics then the mass center of the solar 
system must have a constant velocity relative to it. From what 
has been shown above, it follows that we may change to another 
Newtonian frame in which the mass center of the solar system is 
at rest. This is, in fact, the astronomical frame, to which we 
have referred before. 

If the earth is regarded as a satisfactory Newtonian frame 
of reference, then we must regard as equally satisfactory the 
interior of any vehicle which moves over the earth with constant 
velocity. This is in accordance with common experience: 
we are not conscious of the smooth uniform motion of a train 
when we are traveling in it; we become conscious of the motion 
only when the train lurches or brakes or rounds a corner. 

Frames of reference with translational acceleration. 

Let us now suppose that AS is a Newtonian frame and that *S" 
has relative to it a translational motion with constant accelera- 
tion. Then (cf. Fig. 66), we have 

(5.307) = , rf = ft, 

where o, ft are constants. The relations (5.302) hold in this 
case also, and differentiation gives 

(5.308) -jc = x' + ao, y = y' + ft. 

Let f, f denote, respectively, the accelerations of A relative to 
S and S' 9 and let f denote the acceleration of A" relative to 8; 
then (5.308) may be written 

(5.309) f = f + fo. 

This is called the law of composition of accelerations. 
The equation of motion (5.305) now leads to 

(5.310) tnf - P - mf . ' 

Thus the Newtonian law of motion does not hold relative to S'. 
But we can say that the Newtonian law holds provided that we 
add to the true force P a fictitious force wf . 

As an illustration, consider an elevator descending with constant accelera- 
tion /o. Relative to the elevator, everything takes place as if the elevator 



142 PLANE MECHANICS SBC. 5.3 

were at rest and every particle experienced an upward force mf Q , where m 
is the mass of the particle, in addition to the downward force mg. These 
fictitious forces alter the reactions among the particles constituting the 
human body, and so we are conscious of an acceleration, even though we 
cannot look outside our frame of reference. 

Frames of reference rotating with constant angular velocity. 

Let us now suppose that S is a Newtonian frame of reference 
and S' a frame of reference rotating about a point of S with 
constant angular velocity . Let i, j be perpendicular unit 
vectors, fixed in S' (Fig. 68). Let A be a moving particle. (We 
may think of A as a fly walking on a rotating sheet of cardboard.) 

Taking axes Oxy in S', in the di- 
rections of i and j, the position 
vector of A is 

r - xi + 2/j. 




Fio. 68. Rotating frame of dt dt ' 

reference. 

and so differentiation of (5.311) 
gives, for the velocity of A (relative to S), 

(5.313) q = f = (x - ort/)i + (y + <*c)j. 

Another differentiation gives, for the acceleration of A (relative 
toS), 



(5.314) f = q = (x - 2o> - rfx)\ + (y + 2ax - co 2 i/)j. 

Thus, if X, Y are the components of true force in the directions 
of i, j, respectively, we have the equations of motion 

(5.315) m(x - 2uy - rfx) X, m(y + 2<*x - tfy) = Y. 
These may also be written 

(5.316) mx - X + X' + X", my - Y + Y' + 7", 
where 

X' = 2ma$ Y' = -- 



SEC. 5.3] METHODS OF PLANE DYNAMICS 143 

Thus we may say that the particle moves relative to the rotating 
frame of reference in accordance with Newton's law of motion t 
provided that we add to the true force the two fictitious forces (X 1 , Y') 
and (X", 7"). 

The fictitious force (X', Y') is called the Coriolis force. Its 
magnitude is proportional to the angular velocity of S' and to 
the speed q r of the particle relative to S'; its direction is per- 
pendicular to the velocity q' relative to S', and is obtained from 
the direction of q' by rotation through a right angle in a sense 
opposite to the sense of the angular velocity (Fig. 69). The 
fictitious force (X" y Y") is called the centrifugal force. Its 
magnitude is proportional to the square of the angular velocity 




^S^O) 



mw 2 r 

(Centrifugal force) 



2mwq' 
(Coriolis force) 



FIG. 69. Centrifugal force and Coriolis force in a rotating frame of reference. 

of S' and to the distance of the particle from the center of rota- 
tion; it is directed radially outward from the center of rotation. 

The frame of reference which we employ in ordinary life is the 
earth. It rotates relative to the astronomical frame with an 
angular velocity of 2ir radians per sidereal day; since one sidereal 
day contains 86,164.09 seconds (cf. page 14), the angular velocity 
of the earth is 7.29 X 10~ 6 radians per second. This is a very 
small angular velocity, and hence the Coriolis force and the 
centrifugal force arising from the earth's rotation are not notice- 
able in our daily lives. They are important geographically, 
however; the centrifugal force is responsible for the equatorial 
bulge on the earth, and the Coriolis force is responsible for the 
trade winds. 

When frames of reference turning rapidly relative to the earth 
are employed, these fictitious forces may assume serious pro- 
portions. Thus, in an airplane turning in aerial combat or 
coming out of a dive, centrifugal force may be much greater 
than the force of gravity. 



144 



PLANE MECHANICS 



[SEC. 5.3 



Statical effects of the earth's rotation. 

In Sec. 3.1 we gave an introductory discussion of the force 
of gravity and the weight of a body .near the earth's surface, 
leaving the earth's rotation out of account, i.e., treating the 
earth as a Newtonian frame. We now see that it may indeed be 
so treated provided that the proper fictitious forces are added. 
If we deal only with statical problems, i.e., those in which the 
system is at rest relative to the earth, there is no Coriolis force, 
and so the only fictitious force is centrifugal. The "weight" 
of a particle is the resultant of the force of gravity and the 




FIG. 70. Plumb line on the rotating earth. 

centrifugal force, instead of being merely the force of gravity 
alone. Thus the weight of a particle is proportional to its mass, 
and the theory of Sec. 3.1 is valid provided that we understand by 
mg the weight as just defined. 

We shall now bring our theory still closer to reality by taking 
a more accurate model of the earth. As a first crude approxima- 
tion, the earth may be regarded as a sphere of radius R, where 
R = 3960 miles. More accurately, it is an oblate spheroid with 
an equatorial radius of 3963 miles and a polar radius of 3950 
miles. This is the model which we shall accept for the present 
discussion, and we shall assume that the model rotates about its 
polar axis with constant angular velocity ft. 

In Fig. 70, SN is the earth's axis, A any point on its surface, 
and AB the perpendicular dropped on SN. The gravitational 
attraction of the earth on a particle at A acts along some line 



SBC. 5.3J METHODS OF PLANE DYNAMICS 145 

AC which intersects SN\ its magnitude is proportional to the 
mass m of the particle, and we shall denote it by mg'. The 
centrifugal force is directed along BA, and its magnitude is 
rapQ 2 , where p = BA. 

In order that the particle may remain in equilibrium relative 
to the earth, a third force must be applied to balance the gravita- 
tional force and the centrifugal force. This force must lie in 
the plane ABC, and its magnitude must be proportional to m. 
We denote it by mg] it may be supplied by the tension in a string 
or plumb line, or by the reaction of a smooth plane. We define 
the vertical AV at A as the direction of this force and the hori- 
zontal plane HAH' as the plane perpendicular to it. 

We now ask : As we range over the earth's surface, what is the 
relation between g and g', and what is the inclination of the 
vertical to the direction of the gravitational force? 

The astronomical latitude X is defined as the elevation of the 
astronomical pole above the horizontal plane, i.e., the angle 
between SN and H'AH, or (equivalently) the angle between BA 
and AV. Let us denote by the angle between CA and AV. 
Resolution of forces along and perpendicular to AC gives as 
conditions of equilibrium. 

OS (X - 0), 



Now the ratio pW/g is small; hence 6 is small, and cos 6 differs 
from unity by a small quantity of the second order. To the 
first order of small quantities, we have 



(5.319) g f = g + ptt 2 cos X, 6 = ^ sin X. 

a 

The terms involving 12 are small, and to our order of approxima- 
tion we are entitled to replace p and X by approximate values. 
Now X is approximately equal to the angle BAC, and so 

p = CA cos X, 

approximately. According to a well-known law of hydrostatics, 
the surface of the ocean must be tangent to the horizontal plane ; 
in fact, AV is normal to the surface of our model of the earth. 
Thus, since 6 is small, CA is very nearly normal to this surface, 
and so CA = R approximately, where R is the radius of the 



146 PLANE MECHANICS [SEC. 5.4 

earth in the first crude model. Hence (5.319) may be written 
(5.320) g' = g + RW cos 2 X, 



(5.321) 6 = sin X cos X. 

Thus we can find the gravitational intensity g f in terms of 
measurable quantities, g being measured by means of a pendulum 
(cf. Sec. 6.3). Theoretically, g is given by a measurement of 
the tension in a plumb line, but this is not a practical method. 
At the North and South Poles, g = 983 cm. sec~ 2 ; at the Equator, 
g = 978 cm. sec."" 2 Equation (5.321) gives the deviation of the 
plumb line from the direction of the gravitational force; it 
is a maximum at a latitude of 45. 
Other effects of the earth's rotation will be treated in Sec. 13.5. 

5.4. SUMMARY OF METHODS OF PLANE DYNAMICS 
I. Equations of motion of a particle. 

(5.401) raf = P (vector form); 

(5.402) mx Z, my Y (Cartesian coordinates) ; 

(5.403) mq = P tj ^- = Pn (resolution along tangent and 

p normal) ; 

(5.404) m(f - rtf 2 ) = R, r*& = const, (central force). 

lit Principle of angular momentum for a particle* 

(5.405) h = N, 

where 

h = m(xy yx) = mr*6. 

III. Principle of energy for a particle. 

(5.406) T - W, 
where 

T = img 2 = $m(x z + y 2 ), W = work done; 

(5.407) T + V = E (conservation of energy). 

IV. Principle of linear momentum for a system. 

(5.408) M = F, 



SBC. 5.4] METHODS OF PLANE DYNAMICS 147 

where 

n 

M = V m,q ; 
^T 

(5.409) mf F (motion of mass center). 
V. Principle of angular momentum for_a system. 

(5.410) h**N, 
where 

n 

t(i2/ 2A&), N = moment of external forces. 

(This holds with respect to a fixed point and with respect to the 
mass center.) 

VL Principle of energy for a system. 
'(5.411) T = W, 

where 

n 

J? 7 = 5} Wiflf , IF = work done; 

t i 
(5.412) T + V = E (conservation of energy). 

VII. D'Alembert's principle. 

The reversed effective forces ( m t f) and the real forces 
together give statical equilibrium. 

VIII. Moving frames of reference. 

(i) A frame of reference having a translation with constant 
velocity relative to a Newtonian frame is also Newtonian. 

(ii) A frame of reference having a translation with constant 
acceleration f relative to a Newtonian frame may be treated as 
Newtonian if a fictitious force mf is applied to each particle. 

(iii) A frame of reference rotating with constant angular 
velocity o> relative to a Newtonian frame may be treated as a 
Newtonian frame if to each particle there are applied two 
fictitious forces: 

Coriolis force with components (2mwy, 
Centrifugal force with components (mrfx, 



148 PLANE MECHANICS [Ex. V 



EXERCISES V 

1. At a certain instant, a particle of mass w, moving freely in a vertical 
plane under gravity, is at a height h above the ground and has a speed q. 
Use the principle of energy to find its speed when it strikes the ground. 

2. What is the least number of revolutions per minute of a rotating drum, 
2 feet in internal diameter, in order that a stone placed inside the drum may 
be carried right round? Assume that the contact between the stone and 
the drum is rough enough to prevent sliding. (The reaction of the drum on 
the stone must be directed inward.) 

3. A skier, starting from rest, descends a slope 1 17 yards long and inclined 
at an angle of sin" 1 fa to the horizontal. If the coefficient of friction 
between the skis and the snow is $, find his speed at the bottom of the slope. 
If the skier with his equipment weighs 200 lb., how much energy is dissipated 
in overcoming friction? 

4. A bead of mass m slides on a smooth wire in the form of a parabola 
with axis vertical and vertex downward. If the bead starts from rest 
at an end of the latus rectum (of length 4a), find the speed with which it 
passes through the vertex. Find also the reaction of the wire on the bead 
at this point. 

6. A heavy particle rests on top of a smooth fixed sphere. If it is slightly 
displaced, find the angular distance from the top at which it leaves the 
surface. 

6. Two barges of masses mi, ra 2 at a distance d from each other are 
connected by a cable of negligible weight. One barge is drawn up to the 
other by winding in the cable. If neither barge is anchored, nnd the distance 
through which each barge moves. (Neglect any frictional effects due to the 
water.) 

7. An airplane with an air speed of 120 miles per hour starts from A 
to go to B which is northeast of A. If there is a wind blowing from the 
north at 20 miles per hour, in what direction must the pilot point the air- 
plane if he wishes to go in a straight line from A to 5? 

8. A heavy particle is suspended from a fixed point by a light string of 
length a. If the string would break under a tension equal to twice the 
weight of the particle, find the greatest angular velocity at which the string 
and particle can rotate as a conical pendulum without the string breaking. 

9. A steamer sailing east at 24 knots is 1000 feet to the north of a launch 
which is proceeding north at 7 knots. Find the shortest subsequent dis- 
tance between them if these courses are maintained. Draw rough dia- 
grams, showing 

(i) the tracks relative to the water, 

(ii) the track of the steamer relative to the launch, 

(iii) the track of the launch relative to the steamer. 

10. An automobile travels round a curve of radius r. If h is the height of 
the center of gravity above the ground and 2a the width between the wheels, 
show that it will overturn if the speed exceeds -\/gra/h, assuming no side- 
slipping takes place. 



Ex. V] METHODS OF PLANE DYNAMICS 149 

11. Every second n gas molecules, each of mass m, strike the side of a 
box. Use the principle of linear momentum to find the force required to 
hold the side of the box in place, assuming that each molecule has the same 
speed q before and after hitting the side and that the molecules move at right 
angles to the side. Explain precisely what dynamical "system " you use. 

12. Explain how a man standing on a swing can increase the amplitude 
of the oscillations by crouching and standing up at suitable times. 

13. A light string is attached to a fixed point and carries at its free end 
a particle of mass m. The particle is describing complete revolutions about 
under gravity, and the string is just taut when the particle is vertically 
above 0. Find the tension in the string when in a horizontal position. 

14. Show that, if an airplane of mass M in horizontal flight drops a bomb 
of mass m, the airplane experiences an upward acceleration mg/M. 

16. A chain of any number of links hangs suspended from one end. The 
suspension and the connections between the links are smooth. The chain is 
displaced in a vertical plane and released from rest. Show that in the 
resulting oscillations the center of gravity of the chain never rises higher 
than its initial position, and that if it does ever rise to that same height, 
the whole chain is at rest at that instant. 

16. A wheel spins in a horizontal plane about a vertical axis through its 
center; the bearings are supposed frictionless, and there is a mass clipped on 
one spoke. During the motion the mass slips along the spoke out to the rim. 
Does this cause an increase or decrease in the angular velocity of the wheel? 

17. Assuming that a skier keops his legs and body straight and neglecting 
friction and air resistance, show that he must keep his body perpendicular 
to the slope of a hill in order that he may preserve his balance without 
support from the forward or rear ends of his skis. Give a general discussion 
of the proper direction for his body when friction and air resistance* are 
taken into account. 

18. An ice floe with mass 500,000 tons is near the North Pole, moving west 
at the rate of 5 miles a day. Neglecting the curvature of the earth, find the 
magnitude and direction of the Coriolis force. Express the magnitude in 
tons wt. 

19. A particle moves in a smooth straight horizontal tube which is made 
to rotate with constant angular velocity about a vertical axis which 
intersects the tube. Prove that the distance of the particle from the axis 
is given by 

r = Ae at -f Be'"', 

where A and B are constants depending on the initial position and velocity 
of the particle. 

If when t = the particle is at a distance r a from the axis, what 
velocity must it have along the tube in order that after a very long interval 
of time it may be very close to the axis? 

20. A loop of string is spinning in the form of a circle about a diameter of 
the loop with constant angular velocity. Neglecting gravity, prove that 
the mass per unit length of the string must be proportional to cosec 3 0, 
6 being measured from the diameter. 



160 PLANE MECHANICS [Ex. V 

21. A particle moves with constant relative speed q round the rim of a 
wheel of radius a; the wheel rolls along a fixed straight line with uniform 
velocity F. Taking the wheel as frame of reference, find the Coriolis force 
and the centrifugal force. Indicate them in a diagram. 

22. A system of particles moves in a plane. Prove that, provided the 
mass center is not at rest, there exists at time t a straight line L such that the 
angular momentum about any point on L is zero. Further, show that, if 
no external forces act on the system, the line L is fixed for all values of t. 



CHAPTER VI 

APPLICATIONS IN PLANE DYNAMICS MOTION OF A 

PARTICLE 

6.1. PROJECTILES WITHOUT RESISTANCE 

The science of ballistics is concerned with the motion of projec- 
tiles. The theory of the explosion of the charge and the motion 
of the projectile in the barrel of the gun belong to interior ballis- 
tics, with which we shall not be concerned. After the projectile 
leaves the barrel of the gun, it moves under the influence of 
gravity and the resistance of the air; the purpose of exterior 
ballistics is to predict, from given muzzle velocity and angle of 
elevation of the gun, the path or trajectory of the projectile. 

On account of the complicated nature of the resistance of the 
air, an accurate mathematical prediction is not possible. The 
greatest difficulties arise from the fact that the projectile is of 
finite size. To avoid these, we regard the projectile as a particle. 

In the present section, we shall make a further and much more 
drastic simplification; we shall assume that no resistance is 
offered by the air. We cannot claim that the theory based on 
this hypothesis gives results of much practical value in ballistics, 
except in the case of projectiles thrown with small velocities.* 

The parabolic trajectory. 

Let Oxy be rectangular axes, Ox being horizontal and Oy 
vertical, directed upward. The equations of motion of a particle 
under the influence of gravity are 

(6.101) mx = 0, my = -mg. 
Integration gives 

(6.102) x = u Q , & - VQ gt, 

(6.103) x = XQ + u<>t, y = y Q + vrf ifltf 2 , 

where X Q , yo, u^ VQ are constants of integration. It is evident 

*Cf . C. Cranz and K. Becker, Handbook of Ballistics (H. M. Stationery 
Office, London, 1921), Vol. I, p. 17. 

151 



152 PLANE MECHANICS [Sec. 6.1 

that (a?o, 2/0) is the position and (t*o, VQ) the velocity, both at 
time t = 0. The equations (6.103) give the path, or trajectory, of 
the particle. 

Obviously, at a certain instant (t = v Q /g), we have y = 0, 
so that at that instant the velocity is horizontal. Now the origin 
O and the instant from which t is measured may be chosen as 
we please. Let us choose at the point where the velocity is 
horizontal and measure t from that instant. Then we have 

(6.104) for t = 0, x = y = 0, y = 0, 
Substituting in (6.102) and (6.103), we obtain 

XQ = ?/o = 0, VQ = 0, 

and so the equations of the trajectory become 

(6.105) x = u Q t, y = -ijf. 

Elimination of t gives the equation of the trajectory in the form 

(6.106) = - g, 

a parabola with its vertex at the origin (Fig. 71). 
The focus F of the parabola is situated at a distance a = ^u\]g 

below the vertex, and the directrix L is at the same height above 

the vertex. At any point on the 
trajectory the speed q is given by 

(6.107) 2 = x* + y* = ul + gH* 
= u\ - 2gy = 2g(a - y); 

thus the speed at any point P on the 
trajectory is equal to the speed ac- 
p x quired in free fall to P from rest at 

FIG. 71. Parabolic trajectory, the directrix L. 

with focus F, vertex o, and From this it follows that, whert 

directrix L. , ., . _ , . ' 

a projectile is fired from a point 

P with speed g , the directrix of its parabolic trajectory is 
at the greatest height reached by a second projectile, fired 
straight up from P with the same speed q . In particular, we 
note that all trajectories obtained by firing projectiles at various 
inclinations in one vertical plane, but with a- common initial 
speed g , have a geometrical property in common, namely, a 
common directrix. 



SEC. 6.1] 



MOTION OF A PARTICLE 



153 



The axes shown in Fig. 71 are the simplest for the discussion of 
general properties of the trajectory. But in ballistic problems 
it is preferable to take the origin at 
the point of projection and measure 
the time from the instant of firing 
(Fig. 72). 

If a. is the inclination of the initial 
velocity to the horizontal, we have, 
as in (6.102) and (6.103), 




x = 



cos 



O 



FIG. 72. Parabolic trajectory 
referred to the point of pro- 
jection. 



The projectile strikes the ground when y = 0, that is, when 



(6.109) 

then 

(6.110) 



sin a; 



x = 2P sin 2<x. 
Q 



We note that it is a maximum 



This is the range of the projectile, 
(for given q Q ) when a = 45. 

The greatest height attained by the projectile is obtained by 
putting y = 0. Then, by (6. 108) , 

(6.111) 



t = sin a, 



y = ^ 




Limits of range. 

Let us suppose that a gun gives 
to a projectile a muzzle velocity 
#o- The gun can be pointed in 
any direction. What region in 
space can be reached by the pro- 
jectile? 

The question may be answered analytically, but the most 
elegant solution is geometrical. 

We may confine our attention to one vertical plane through 
the gun, which is at in Fig. 73. First we ask: Where is the 
focus of the trajectory passing through an assigned point P? 
The answer is given by the following construction: 



Co 

FIG. 73. Construction for the 
foci of the two parabolic trajectories 
passing through P. 



154 



PLANE MECHANICS 



[SEC. 6.2 




FIG. 74. Paraboloidal region within 
ran go. 



Draw the directrix L, at a height %ql/g above 0. Draw the 
circle C with center 0, touching L at A. Since is a point on 
the parabolic trajectory, the focus must lie at a distance OA 

from 0, and so it must lie on 
Co. Similarly, if the circle C 
is drawn with center P to 
touch L, the focus must lie 
on C also. Hence the focus 
must lie at an intersection of 
the circles Co and C. In gen- 
eral, there will be either two 
points of intersection, FI and 
Ft, or none. In the former 
case, P is within range and F\, 
Fz are the foci of the two trajectories through it. In the latter 
case, P is out of range. 

If P is at the limit of range, the circles C and C touch. Then 
P is equidistant from and a horizontal line LI, drawn at a height 
twice that of L, i.e., at a height q\/g. Thus the locus of P in 
space is a paraboloid of revolution, having for focus and A for 
vertex (Fig. 74). All points inside this paraboloid are within 
range, and all points outside it are out of range. 

6.2. PROJECTILES WITH RESISTANCE 
General equations. 

We turn now to the more practical problem in which the air 
exerts on the projectile (still regarded as a particle) a force R 

acting in a direction opposite to 
the velocity (Fig. 75). 

In resolving forces and accel- 
eration in order to obtain equa- 
tions of motion in scalar form, 
we have a choice of two pro- 
cedures: (i) resolution along 
horizontal and vertical direc- 
tions, and (ii) resolution along 
the tangent and normal to the 




R 



mg 



FIG. 75. The forces acting on a 
projectile. 



trajectory as in (5.103). 

Denoting by 6 the inclination to the horizontal of the tangent 
to the trajectory, we obtain by (i) the equations 



SBC. 6.2] MOTION OF A PARTICLE 155 

(6.201) mx ~ -R cos 0, my = R sin - mg. 

Denoting by p the radius of curvature of the trajectory, we 
obtain by (ii) the equations 

(6.202) mq ^ = -R - mg sin 0, ^ = wgr cos 0, 

where q is the speed and ds an element of arc of the trajectory. 
The equations (6.202) are, of course, only a different mathe- 
matical expression of (6.201). 

The resistance experienced by a given projectile depends on its 
speed and on the density of the air. Regarding the air as 
stratified into horizontal layers each of constant density, so that 
the density is a function of y only, we may express R in the form 

(6.203) R = R(y, q) t 

to show that it is a function of y and q only. 
Since 

(6.204) cos 6 = -; sin = t 

q Q 

we may write (6.201) in the form 

(6.205) = -$.r, y = -$y - g, 
where < is a function of y, q, namely, 

(6.206) *(y, q) = 

The mathematical problem of the determination of the 
trajectory is made much more difficult by the fact that there is no 
physically valid formula expressing R as a function of q. For 
small values of q, R varies as q] but this simple law breaks down 
before we reach those velocities which are of interest in ballistics. 
For low ballistic velocities, R varies as q 2 ; but this law again 
breaks down when the velocity of the projectile approaches the 
velocity of sound. The law of dependence is then complicated 
and can be represented only graphically or by tables of values 
obtained experimentally. Hence, we must not expect to find any 
simple formulas for trajectories. In general, (6.205) must be 
integrated by a tedious process of step-by-step numerical integra- 
tion. The differential equations are integrated approximately 
over small intervals of time; the errors due to approximation 
become insignificant when the intervals are very small. 



156 



PLANE MECHANICS 



[SBC. 6.2 



The difficulties involved in the above method lead us to do 
what we so often do in applied mathematics replace the com- 
plicated physical problem by one that is simpler mathematically. 
Thus, we shall confine our attention below to the case where R is 
independent of y and devote particular attention to the cases 
where R varies as g 2 or as q. 

Resistance independent of height. 

If the resistance depends on the speed only, so that R = R (#), 
the problem is most easily attacked by means of (6.202). Let us 
write 

(6.207) R = mg<t>(q) 

for convenience. Noting that decreases as the arc length s 
increases, we have p = ds/dOj and elimination of ds from 
(6.202) gives 

ldg = </>(<?)+ sin 0. 

q dd 



(6.208) 

v ' 



(6.209) 



dx 

~TZ 
dd 

dy 

-^ 
dd 



, . 

~TZ = ~J~ TB = ^~P cos = -~ ' 



y y s . n q an 

-^ = -J^ -JZ = p sm = % --- y 



cos 6 

This is called the equation of the hodograph, since #, 6 are the polar 
coordinates of a point on the hodograph. 

If we can solve x (6.208), all desired information about the 
trajectory may be obtained by quadratures. By (6.202), we 
have 

* ' 9l 

g 

# 2 jtan 

g 

_ *? sec 

q g 

Let us suppose that the projectile is fired from the origin at time 
t = with speed q Q at an angle of elevation . Let 

(6.210) q = /(0) 

be the solution of (6.208), supposed known. Then integration of 

(6.209) gives 



(6.211) 



_ 
dd 



dx ds 
~dsd6 
dy cte 
3* dd 
dtds 
dsdO 



Bf(B)dB. 



These equations express x, y, t as functions of one parameter 
and so determine the trajectory. 



SEC. 6.2] MOTION OF A PARTICLE 157 

But we cannot use (6.211) until the function /(0) is known, 
i.e., until the differential equation (6.208) is integrated. If the 
function <f>(q) is general, the integration of (6.208) cannot even be 
reduced to quadratures. For some special forms of <f)(q) the 
integration can be reduced to quadratures, and in some cases the 
solution /(0) can be expressed in terms of elementary functions. 
We shall consider bolow the case <t>(q) = C# 2 , but before making 
this special choice of < we shall transform (6.208) by changing 
to a new independent variable $, defined by 

tanh ^ = sin 0. 
It is easily seen that (6.208) transforms into 

(6.212) i = tanh * 



Resistance varying as the square of the velocity. 

Let us take the law of resistance to be 

(6.213) R = m<7<K<?), 4>(q) = Cq\ 

where C is a constant. Division of (6.212) by %q* gives 



(6.214) = - tanh * ~ 2C > 
^ } dt\q 2 / q 2 

this is a standard type of equation, with solution 

(6.215) ~ = sech 2 ^ (A - 2C J cosh 2 ^ dtf 

= cos 2 e [A - C tanh- 1 (sin 0)] - C sin 0, 

where A is a constant of integration, to be fixed by the initial 
conditions. 

Theoretically at least, the equations (6.211) now determine 
the trajectory, /(0) being the reciprocal of the square root of the 
right-hand side of (6.215). But it is evident that the calculations 
involved are very complicated. 

When the projectile moves in a vertical line, the problem 
is much simpler. If i is a unit vector directed vertically upward 
and the position vector of the projectile is t/i, the acceleration is 
i/i. The force of gravity is mgi. The resistance is mgCy 2 i 
for motion upward (y > 0) and mgCy 2 i for motion downward 



158 PLANE MECHANICS [SBC. 6.2 

(y < 0). Hence the equation of motion is 

(6.216a) g = gCy 2 g for motion upward; 

(6.2166) g =* gCy 2 g for motion downward. 

Since 



we have 

(6.217a) 1 ^ ^., g dy for motion upward; 

l + 



(6.2176) . 2 = g dy for motion downward. 

Thus, on integration, 

(6.218a) y = - log (1 + C# 2 ) + A for motion upward, 



(6.2186) y = ~ log (1 - Cy 2 ) + A' for motion downward, 



where A, A 1 are constants of integration. 

Let us consider two special cases, corresponding to (a) a shell fired verti- 
cally upward, and (b) a bomb dropped vertically downward. 

(a) Let g be the initial velocity of the shell, fired from y = 0. Then, 
by (6.218a), we have 



The height h to which the shell rises Is found by putting y 0; thus, 

(6.220) h = ^ log (1 + CflD. 
If the resistance is small, this gives, approximately, 

(6.221) /i = *~-i^l 

Q Q 

in which the first term is the well-known expression for height attained under 
no resistance. 

(6) Let the bomb be dropped from y with no velocity. Then 
(6.2186) gives 

(6.222) y 2^1og(l - W 

y being, of course, negative. When the bomb has dropped a distance h, 
we have 

log (1 - CM - - 



SBC. 6.3] MOTION OF A PARTICLE 159 

and hence its speed is 

(6.223) q . yj l "*"***- 

As k tends to infinity, g tends to C~*. This is the limiting velocity; its value is 

(6.224) C~* - q Vw/R, 
where R is the resistance at any speed q. 

Relations connecting y and t may be obtained from (6.216) by integration. 
Thus, (6.216a) may be written 



One integration gives y in terms of t, and a second integration gives y. 
Resistance varying directly as the velocity. 

Although the law of resistance 
(6.225) R = mgCq 

is not accurate physically, it is so simple to treat mathematically 
that it is a useful approximation at least an improvement over 
the assumption of no resistance at all. 

Turning back to (6.206), we note that < is now a constant 
($ = gC), and the equations (6.205) are easy to handle, because 
the variables x and y arc separated. We obtain, on integration, 



(6.226) 



where x , y<> are the coordinates and UQ, V Q the components of 
velocity for t = 0. 

When 3> is small, these equations yield approximately 

{x = x + u t i$u 2 , 
y = y. + Vot - fa* - 

in which the terms independent of 3> correspond to the parabolic 
trajectory. 

6.3. HARMONIC OSCILLATORS 

The simple pendulum. 

A simple pendulum consists of a heavy particle attached 
to one end of a light rod or inextensible string, the other end 
of the rod or string being attached to a fixed point. We con- 



160 PLANE MECHANICS [SBC. 6.3 

aider only motions of the pendulum in which the string remains 
in a definite vertical plane. In Fig. 76, B is the point of attach- 
ment and A is the particle (of mass ra), drawn aside from its 
position of equilibrium 0. Oxy are rec- 
tangular axes in the plane of motion, Ox 
being horizontal. 

The particle moves under the. influ- 
ence of two forces: (i) its weight mg, and 
(ii) the tension S in the string. Since S 
acts along the normal to the circular path 
of the particle, it is clear that the dy- 
namical problem presented by the simple 
pendulum is precisely the same as that of 
_ __ the motion of a particle on a smooth 
'* circular supporting curve, fixed in a ver- 

tical plane. 
rmg Let AB = Z, OSA = B] then 

FIG. 76. The simple pen- , Qfm fi _ I y . fi X 

' 



fi _ . fi 

dulum. . COS -- j ') Sin = j> 

where x, y are the coordinates of A. The equations of motion 
are 

(6.302) mx = S sin 0, my = S cos 6 - mg. 

Let us investigate small oscillations about the position of 
equilibrium, assuming x and its derivatives to be small. Then 
y and its derivatives are small, of the second order, and cos 6 
differs from unity by a small quantity of the second order. 
Thus, to the first order inclusive, the second of (6.302) gives 

(6.303) S = mg, 

and substitution from this and from (6.301) in (6.302) gives 

(6.304) x + p*x = 0, p = J2. 

The general solution of this differential equation is 

(6.305) x = A cos pt -h B sin pt, 

where A, B are constants, to be determined by the initial condi- 
tions. A motion given by an equation of this form is called 
simple harmonic. 

Differentiation of (6.305) gives 

(6.306) x = -Ap sin pt + Bp cos pt. 



SBC. 6.3] MOTION OF A PARTICLE 161 

We note that x and x have the same pair of values at times 

ti, ti + r, h + 2r, i + 3r, 
where h is arbitrary, and 

(6.307) r = ^ = 

P 

When the position and velocity of a particle are repeated over 
and over again at equal intervals of time, we say that the motion 
is periodic; the interval is called its periodic time. Hence the 
above motion of a simple pendulum is periodic, with periodic 
time given by (6.307). 

In Sec. 13.2 the finite oscillations of a pendulum will be 
discussed, and it will be found that the finite oscillations are 
periodic but not simple harmonic; the formula for the periodic 
time is different. 

The harmonic oscillator. 

The simple pendulum, executing small oscillations, is only 
one physical example of a type of dynamical system of frequent 
occurrence. Many problems of oscillation can be discussed 
in a single mathematical form; so we create a single mathematical 
model for them all. This model 2 

is called the harmonic oscillator , ( ^ , x 

(Fig. 77). 

Jf, / . .,, , . , FIG. 77. The harmonic oscillator. 

The harmonic oscillator consists 

of a particle which can move on a straight line, which we shall 
take for z-axis. It is attracted toward the origin by a controlling 
force varying as the distance. If i is a unit vector in the positive 
direction of the z-axis, the controlling force may be written 
mp 2 ai, where m is the mass of the particle and p is a constant. 
The acceleration is ai, and hence the equation of motion is 

(6.308) mxi = mp 2 xi, 
or, in scalar form, 

(6.309) * + P 2 x = 0. 

The general solution of this equation may be written in the form 
(6.305), that is, 

(6.310) x = A cos pt + B sin pt, 
so that the motion is simple harmonic. 



162 PLANE MECHANICS [Sue. 6.3 

Let us now define constants a, e by the equations 

(6.311) a = V^ 2 + B\ 

A . B 

cos = j sin e = 



VA* + B* 
Then (6.310) may be written 

(6.312) x = a cos (pt + e). 

We observe that x covers the range (a, a) and that the motion 
is periodic with periodic time 2ir/p. The following terminology 
is used in connection with the harmonic oscillator: 

amplitude = a, 

period or periodic time = T = 2fr/p, 

frequency = number of oscillations per unit time 

= 1/r = p/27r, 
phase = pt + c. 

Effect of a disturbing force. 

Let us now suppose that, in addition to the controlling force, 
there acts on the harmonic oscillator a force whose component 
in the positive direction of the ar-axis is mX. The equation of 
motion (6.308) is now modified to the form 

(6.313) rnxi = - mp*xi + mXi, 
or, in scalar form, 

(6.314) * + p*x = X. 

First let us suppose that X is constant. The general solution 
of (6.314) is then 

(6.315) x = ~ + a cos (pt + ). 

This motion is simple harmonic, but the center of the oscillations 
is displaced to the position x = X/p 2 , as is seen by writing 

(6.315) in the form 

(6.316) x - ~ = a cos (pt + e). 

When the oscillation or vibration has a high frequency, so that p 
is large, the displacement of the center is small. 



SBC. 6.3] MOTION OF A PARTICLE 163 

Let us now suppose that X is itself simple harmonic, varying 
according to the formula 

(6.317) X = k cos ct, 

where k, c are constants. Then the equation of motion (6.314) 
reads 

(6.318) x + p*x = k cos ct. 

In the general case where c j p, the solution is 

k 

(6.319) x = a cos (pt + e) + -, Tz cos c *> 

p c 

where a, e are constants, to be determined by the initial condi- 
tions. Thus the motion is a superposition of an undisturbed 
simple harmonic motion and a second motion; the latter has the 
same period as the force, and an amplitude which becomes large 
when the difference between the periods of the free oscillator and 
the disturbing force becomes small. The great increase in the 
amplitude of the oscillations under this last condition is called 
resonance. We shall discuss it again below, taking resistance 
into consideration. 

Damped oscillations. 

Let us now suppose that, in addition to the controlling force, 
there acts on the particle of a harmonic oscillator a force of 
resistance proportional to the velocity, called the damping force. 
The component of this force in the positive direction of the z-axis 
may be written 2rapt:r, where \L is a positive constant. The 
equation of motion (6.308) is modified to 

(6.320) inn = - rnp' 2 xi 2mnxi, 
or, in scalar form, 

(6.321) x + 2fjix + p*x = 0. 

Now, 

(6.322) x = Ce nt 

will satisfy this equation provided that n satisfies the character- 
istic equation 

(6.323) n* + 2/m + P 2 = 0. 



164 PLANE MECHANICS [Sac. 6.3 

This equation has two roots, which may be real or complex; 
they are 

(6.324) ni, n 2 = -/* VV - P 2 - 
Let us define a real positive number / by 

(6.325) I = V|M 2 ~ P 2 |, 

the sign | | indicating absolute value. We have then two cases 
to discuss: (i) light damping, ju < p; (ii) heavy damping, ^ > p. 

CASE (i) Li^fa damping (/* < p). 

In this case, 

(6.326) ni, n 2 = /* #, 
and the general solution of (6.321) is 

(6.327) x = <7ie<-"+' I)< + tV""- '. 
This may be written 

(6.328) x = <r*(A cos ft + B sin tt), 
where 

(6.329) A = Ci + C 2 , = z(Ci - C 2 ). 

Since, for physical reasons, we are interested only in real values 
of x y it is evident that, although the differential equation is 
satisfied by (6.327) with Ci, C 2 complex constants, we should 
adopt as our final solution (6.328) with real constants A, B. 
These may be chosen to satisfy the initial conditions, i.e., to give 
assigned values to x and x f or t = 0. 
The motion given by (6.328) may also be written 

(6.330) x = a*-*' cos (It 4- c), 

where a, e are constants chosen to fit the initial conditions. 
This is not a simple harmonic motion, not being of the form 
(6.312). The factor e~ M * indicates a general decay of the oscilla- 
tions, x tending to zero as t tends to infinity. 

By changing the instant from which t is measured, we can get 
rid of e in (6.330), obtaining the simpler expression 



(6.331) x = oerM* cos U. 



SEC. 6.3] 



MOTION OF A PARTICLE 



165 



On plotting a; as a function of t, we obtain the graph shown in 
Fig. 78. (The graph of the simple harmonic motion 

x = a cos It 

is a cosine curve, resembling the above curve, except for the 
latter's tendency to die away.) The curve of (6.331) should be 
compared with the curves 

(6.332a) 
(6.3326) 




FIG. 78. Position-time graph for a lightly damped harmonic oscillator. 

also shown in Fig. 78. It is easily seen that (6.331) touches 
(6.332a) and (6.3326) at 



(6.333a) 
(6.3336) 



I ' 

(2nH 



I 



respectively, n being any integer. 

We may regard (6.331) as a harmonic motion with decaying 
amplitude, given by ae~ M< . But the following is a more accurate 
description. 

The maxima and minima of (6.331) occur at instants t t n , 
where 



(6.334) 



lt n = n-K a, 



tan a. 



V 



166 PLANE MECHANICS fSBC. 6.3 

n being any integer. The common interval is 

(6.335) r = ; 

thus 2ir/l may be referred to as the period of the oscillations. 
The values of x corresponding to (6.334) are 

(6.336) x n = ( l) n ae-^ cos a. 
Thus successive values are connected by 

(6.337) = -e-"+ 1 -< ) = -cr**". 

#n 

The ratio of successive swings to opposite sides is, in absolute 
value, 

(6.338) Xn+l 



It is usual to define the logarithmic decrement of the damped 
oscillation as the logarithm of the reciprocal of this ratio to base 
10; its value is (TT/Z/Q logic e. 

CASE (ii) Heavy damping (/* > p). 

In the case of heavy damping, we have, by (6.324), 

(6.339) rci, ri 2 = -M + I, 
and so the general solution of (6.321) is 

(6.340) x = e-^(Ac lt 4- Be~ lt ). 
Then, 

(6.341) x = er[A(l - n)e - B(l 
This vanishes only if 



Since e 2 " is a steadily increasing function of t, there can be at 
most one solution. Hence the velocity of the oscillator vanishes 
at one instant at most. The motion is non-oscillatory, or 
deadbeat, x tending to zero as t tends to infinity, since /u > L 

Forced oscillations. 

Finally, let us suppose that a harmonic oscillator is subject to 
(i) a controlling force ( mp z x), 
(ii) a damping force ( 2mp,x), 
(iii) a disturbing force (mk cos ct). 



SBC. 6.3] MOTION OF A PARTICLE 167 

The equation of motion is now 

(6.343) x + 2t*x + p*x k cos ct. 
The general solution is 

(6.344) x = x l + x, } 
where x\ satisfies 

(6.345) x, + 2 M zi + p'zi = 0, 

and contains two constants of integration, while x 2 is any partic- 
ular solution of 

(6.346) x 2 4- 2^2 + p*x t = k cos ct. 

Then x\ corresponds to the motion of the damped oscillator 
without disturbing force; it is given by (6.330) or (6.340), 
according as the damping is light or heavy. 

As for 2 , (6.346) is satisfied by the expression 

(6.347) x* = E cos ct + F sin ct, 

where 

. _ k(p 2 c 2 ) 

(b.d4S) 4 - ( p2 _ C 2)2 

This solution is most easily found by replacing cos ct by e lct 
in (6.346) and finding a complex constant G so that Ge lct is a 
solution; x% will then be the real part of this solution. Alter- 
natively, we may substitute (6.347) directly in (6.346) to find the 
constants. 
If we define 



_ , + (2MC) 

2 __ C 2 

00817 



we may write (6.347) in the form 

(6.350) * 2 = b cos (d 4- rj). 

Thus the general motion of the damped oscillator under the 
influence of the disturbing force is 

(6.351) x xi + 6 cos (ct + 77). 



168 PLANE MECHANICS (Sfic. 6.4 

As $> oo, zi-0; thus after a long time the motion tends 
to the forced oscillation 

(6.352) x = b cos (ct + r?). 

This is a simple harmonic motion of amplitude 6 and period 
equal to that of the disturbing force, but there is a difference 
in phase. 

If the period of the disturbing force is equal to the free period 
of the oscillator, we have the case of resonance. Then c = p, 
77 = ^?r, and so 

k k 

(6.353) x = cos (pt - fr) = sin pt, 



the disturbing force being mk cos pt. The difference of TT in 
phase is interesting. 

6.4. GENERAL MOTION UNDER A CENTRAL FORCE 

Cartesian equations and the law of direct distance. 

Consider a particle of mass m attracted toward a fixed point 
by a force mP. (We may include the case of repulsion by taking 
P negative.) For rectangular Cartesian coordinates Oxy, the 
equations of motion are 

, AM \ . mPx . mPy 

(6.401) mx = -- , my = -- -> 

where r 2 = x 2 + y 2 . We refer to this as motion under a central 
force, because the line of action of the force passes through a 
fixed center 0. 

As an application of (6.401), let us consider the law of direct 
distance, meaning thereby that P is proportional to r. Let us 
confine our attention to an attractive force, putting 

(6.402) P = jfcr, 

where & is a constant. 
Then (6.401) read 

(6.403) x + k z x = 0, y + k*y = 0; 
these equations have the general solutions 

( x = A cos kt + B sin kt, 
\y = Ccoskt + D sin kt, 



SBC. 6.4] MOTION OF A PARTICLE 169 

where the coefficients are constants, to be fixed by the initial 
conditions. 

If we solve (6.404) for cos kt and sin kt, and eliminate t by 
the identity 

cos 2 kt + sin 2 kt = 1, 
we get 

(6.405) (Cx - Ay) 2 + (Dx - By) 2 = (BC - AD)\ 

This is a central conic and necessarily an ellipse since x, y remain 
finite, as we see from (6.404). Thus the orbit described under a 
central attractive force varying directly as the distance is an ellipse 
having its center at the center of force. This motion is called 
elliptic harmonic. 

To illustrate the significance of the constants in (6.404), 
let us suppose that at time t the particle is at x a, y = 0, 
moving with velocity VQ in the direction of the t/-axis, so that 
x = 0, y = VQ. Putting this information into the equations 
(6.404), first as they stand and then in the form obtained by 
differentiation, we get 

a = A, = 0, 

= , V Q = Dk, 

and so the motion is given by 

(6.406) x = a cos kt, 2/ = T~ sm ^- 

K 

Polar coordinates. 

Returning to the general problem of motion under a central 
force, we shall now obtain equations easier to handle than 
(6.401). It is only in the case of the law of direct distance that 
(6.401) are convenient. 

Since the force on the particle passes through 0, it has no 
moment about 0. Hence, by the principle of angular momentum 
(5.214), the angular momentum about is constant. We shall 
change slightly the notation of Chap. V, now letting h denote 
angular momentum per unit mass; then, by (5.106), 

(6.407) h = r*6 = constant, 

where r, are the polar coordinates of the particle. An alter- 



170 PLANE MECHANICS [SBC. 6.4 

native expression for h is pq, where p is the perpendicular from 
on the velocity vector q. 

If we follow the radius vector, drawn from to the particle, 
we observe that when it turns through an infinitesimal angle d6 
it sweeps out an area ^r 2 d&. Thus the arcal velocity A may be 
defined as 

(6.408) A = r 2 0, 

A being, in fact, the total area swept out from some initial 
instant. We observe that h is twice the arcal velocity, and by 
(6.407) we have the following important result: In motion under a 
central force, the arcal velocity is constant. 

This fact is us&l to simplify the problem of determining the 
orbit. We define 

(6.409) u = ~, 

the reciprocal of the radius vector. Then, since (6.407) may be 
written 

(6.410) 6 = hu*, 

we have 

1 du A , du 



(6.411) 



The vector equation of motion is 

(6.412) mi = mP, 

where P is the attractive force per unit mass. Resolution along 
the radius vector gives, by (4.107), 

(6.413) f - r6* = -P, 

where P is the inward component of P. By (6.411), we obtain 

(6.414) gf -f 



SBC. 6.4] MOTION OF A PARTICLE 171 

This is the differential equation of the orbit of a particle moving 
under an attractive central force P per unit mass. If we can solve 
this equation, obtaining u as a function of 0, we have the equation 
of the orbit in polar coordinates. 

Henceforth, let us suppose that P is a function bf r only. It 
is easy to see that the work done in passing from one position to 
another is then independent of the path described. Thus the 
system is conservative, and we may use the principle of energy. 
Let T, V, E be respectively the kinetic energy, potential energy, 
and constant total energy, all per unit mass. Then 

(6.415) T + V - E. 
Now, by (6.411), 

(6.416) T = i(f* + r*) - \ 

The potential energy per unit mass is such that 

P = - grad F, 
so that, since P is the inward component of P, 



(6.417) p = , v 

where ro is some constant. Hence, (6.415) gives 



This equation is really equivalent to (6.414), as may be seen on 
differentiation. We are, of course, to remember that V, being a 
function of r, is also a function of u. 

Apsides and apsidal angles. 

An apse is a point on an orbit at a maximum or minimum 
distance from the center of force. The condition for an apse is 
r = 0, or, equivalently, 

(6.419) *f - 0. 

From (6.418) it follows that, at an apse, 






172 PLANE MECHANICS [SEC. 6.4 

Since V is a function of u, this is an equation to determine the 
values of u at the apsides, supposing the constants E and h 
known. 

As an illustration, let us return to the case P = k 2 r. Then 

(6.421) F = P 2 r 2 = -5, 
and so (6.420) may be written 

(6.422) w 4 - p (#^ 2 - P 2 ) = 0. 

This quadratic equation in u 2 yields roots wf, u\, which will 
be the squares of the reciprocals of the semiaxes of the elliptical 
orbit. 

By a study of apsides, we may obtain a general description of 
an orbit without actually solving the differential equation (6.414). 
To establish an important feature, let us temporarily forget the 
dynamical problem and think of a differential equation 

(6.423) ^ = f(y), 

to be solved under the initial conditions y = i/o, dy/dx for 

x = 0. These initial conditions de- 
termine a unique solution. Since the 
transformation x x' leaves the 
form of the differential equation and 
the initial conditions unaltered, it is 
evident that the curve y = F(x) y which 
satisfies (6.423) and the initial condi- 
^ tions, is symmetric with respect to the 




Since P is a function of u, it is tjlear 
that (6.414) and (6.423) are equations 

Fm. 79.-Symmetry of a of the same f orm > with the COrrespond- 
central orbit with respect to an enCC U*y,6*X. If W6 measure 6 

apse me. from an apse, the initial conditions 

for (6.414) are the same as those for (6.423). Hence, a central 
orbit is symmetric with respect to the line drawn from the force 
center to an apse. 

This result throws much light on the general structure of a 
central orbit. Let A and B (Fig. 79) be consecutive apsides, and 



SEC. 6.4] MOTION OF A PARTICLE 173 

let us suppose that the portion AB of the orbit is known. The 
orbit is symmetrical about OB; hence, we may obtain some more 
of the orbit by folding the portion A B over the line OB, obtaining 
BC, with an apse at by symmetry. The apsidal distance 00 
is equal to the apsidal distance OA. Again, folding BC over 00, 
we get CD with an apse at D, and OD = OB. 

The following facts are now clear: 

(i) Any central orbit has only two apsidal distances. The 
orbit is a curve touching two concentric circles, the radii of which 
are the two apsidal distances. 

(ii) Once the orbit between two consecutive apsides is known, 
the whole of the orbit may be constructed by operations of fold- 
ing over apsidal radii. 

(iii) The angle subtended at the center by the arc joining 
consecutive apsides is a constant. It is called the apsidal 
angle. 

In some cases, (i) may be violated. The radius of the inner 
circle may be zero, or that of the outer circle may be infinite. 
But these are to be regarded as exceptional cases. 

Let us now consider how the apsidal angle is to be found. 
When u is increasing, (6.418) gives 

(6.424) ^ 
where 

(6.425) F(u) - 
Thus 

d = _*L 



and so, if MI, u z are the reciprocals of the apsidal distances (with 
Ui < u%), the apsidal angle a is 

(6.426) 



As an illustration, let us consider the law of direct distance, 
where F is given by (6.421). Here F(u) is of the form 



174 PLANE MECHANICS [Sue. 6.4 

where A and B are constants. But F(u) = at an apse; hence 
u = 1*1, u = u z are roots of F(u) = 0, and so 



F( u) 
Thus, by (6.426), 



as we already knew from the fact that the orbit is a central 
ellipse. 

Stability of circular orbits. 

In a circular orbit u is constant. Hence by (6.414) the 
possible radii of circular orbits are determined by 

(6.428) = I- 

With an attractive force (P > 0), we can obtain a circular orbit 
of any radius, by projecting the particle at right angles to the 
radius vector with that velocity which makes 

(6.429) h* 

VI 

But the question arises: Are these circular orbits stable or 
unstable f In other words, if slightly disturbed, will the resulting 
orbit lie close to the original circular orbit, or will it deviate far 
from it? This question is important physically, because in 
nature small disturbances are always present, and they will 
destroy an unstable circular orbit. The only circular orbits we 
r*n hope to observe are those that are stable. 

It should be clearly understood that we shall not consider the 
effects of disturbing forces which continue to act on the particle; 
we assume that the position and velocity of the particle have 
been disturbed and investigate the resulting motion under the 
original central force. 

Let u = u Q and h = h Q in the circular orbit. Then, by (6.429), 

(6.430) hi : 



To study the disturbance, we put 
(6.431) u = -Uo + {, 



SEC. 6.4) MOTION OF A PARTICLE 175 

where J and its derivatives are assumed to be small. We also 
assume that h ho is small. Substitution in (6.414) gives 

(6.432) + .. + 



Now, on expansion in powers of {, 

<6 - 433 > Ffsrl - K, (' + &" 



where P' = dP/du, and the subscript indicates evaluation 
for u = w . 

Then, to the first order in small quantities, (6.432) becomes 

(6.434) g + At- - , 
where 

(6.435) A - 1 - ^ (~ J - -) = 3 - ?*, 

hlu* \P UQ/ Po 

by (6.430) ; B is another constant whose value does not interest 
us. The solution of (6.434) is 

(6.436a) J = ~ + Ci cos (\/3 (?) + C 2 sin (\/I 0), 

(6.4366) - + Ci cosh (v^^Z ^) + C 2 sinh (x/ 11 ^ * 0), 
(6.436c) f - i^^ 2 + CJ + C 2 , 

according as A > 0, A < 0, or A =0. 

Of these solutions, only (6.436a) remains permanently small; 
the others increase indefinitely with 0. Hence the circular orbit of 
radius I/UQ is stable if, and only if, 

(6.437) 

In particular, let us consider the case of a force varying 
inversely as the nth power of the distance so that 

(6.438) P = ^ = ku\ 

Then 

uP' 



176 PLANE MECHANICS [Sue. 6.5 

and so circular orbits under the attractive force (6.438) are stable 
if, and only if, 

(6.439) n < 3. 

Thus the law of direct distance (n = 1) and the law of the 
inverse square (n = 2) give stable circular orbits; the law of 
the inverse cube (n = 3) gives unstable circular orbits. 

6.5. PLANETARY ORBITS 
The law of the inverse square. 

As already remarked in Sec. 3.1, Newton's law of gravitational 
attraction states that two particles of masses w, w', at a distance 
r apart, attract one another with equal and opposite forces of 
magnitude 

(6.501) 



where G is the gravitational constant. 

Coulomb's law of electrostatic attraction states that two 
particles carrying electric charges e, e' (in electrostatic units), 
at a distance r apart, repel one another with equal and opposite 
forces of magnitude 

PP' 
(6.502) ^. 

If e and e' have opposite signs, this force is a force of attraction. 

Here we have two examples of the law of the inverse square. 
The law (6.501) governs astronomical phenomena in particular, 
the motion of a planet round the sun. The law (6.502) governs 
atomic phenomena in particular, the motion of an electron in 
an atom about the central nucleus. In this case, of course, the 
charges e, e' have opposite signs, so that the force is one of 
attraction, as in the gravitational case. It is remarkable that 
the same form for the law of attraction should hold on such 
different scales. 

The expressions (6.501) and (6.502), combined with Newton's 
law of motion, constitute two hypotheses regarding phenomena 
in gravitational and electrostatic fields. For a long time, they 
were accepted as completely valid from a physical point of view, 
but that is no longer the case. The modern astronomer knows 



Sue. 6.5] MOTION OF A PARTICLE 177 

that gravitational attraction should be discussed in terms of the 
general theory of relativity, and the physicist insists that prob- 
lems on the atomic scale belong to quantum mechanics. It 
would, however, create a completely false impression if we were 
to say that the law of the inverse square has disappeared from 
modern science. Nearly all the calculations of astronomers 
are still based on (6.501) and give results in excellent agreement 
with observation. Moreover, the physicist often falls back on 
the simple atomic picture based on (6.502) and Newton's law 
of motion. 

In what follows, we shall discuss the motion of a planet 
attracted by the sun. Obviously, by a mere change of constant, 
the same reasoning will apply to the motion of an electron in 
an atom. 

Determination of the orbit. 

The sun and a planet are rcgaidcd as particles, of masses 
M and m, respectively. The attraction of the sun on the planet, 
given by (6.501), produces an acceleration GM/r 2 ] and the 
attraction of the planet on the sun produces an acceleration 
Gm/r 2 . These acceleration?* are in the ratio M/m, which is 
actually a very large number. Hence, without any serious 
departure from reality, we may neglect the acceleration of the 
sun and treat it as if it were at rest. Later we shall see how to 
treat the problem exactly. 

We consider then the case of a particle attracted toward a 
fixed center by a force P per unit mass, where 

(6.503) P = 

/* being some positive constant. The differential equation 
(6.414) for the orbit now reads 

(6.504) + - fr 
The general solution is 

(6.505) u = ^ + C cos (0 - ), 

where C and 0o are constants of integration. This is, in polar 
coordinates, the equation of the most general orbit described under 
a central force varying as the inverse square of the distance. 



178 PLANE MECHANICS [SEC. 6.5 

The potential energy per unit mass is 

(6.506) F = JPcJr = -= -/m, 

the constant of integration being chosen to make V vanish at 
infinity. 

Let us now substitute from (6.505) in (6.418), the equation 
of energy, in order to express the constant C in terms of E and h 
(the total energy and angular momentum per unit mass^. We 
get 

C f + j| + 2C^ 2 cos (9 - ) = ~ [# + Jj + nC cos (6 - 
so that 

(6.507) C> = g + 

By rotating the base line = 0, we can make = and 
C > in (6.505) ; this we shall suppose done. Then the equation 
(6.505) for the orbit reads 



e \ 



(6.508) u = p l + -l+r cos e 

From the focus-directrix property of a conic, we know that 
its equation in polar coordinates may be written 

(6.509) u = i (1 + e cos 0), 

where I is the semi-latus-rectum (i.e., half the focal chord parallel 
to the directrix) and e the eccentricity; 6 is measured from the 
perpendicular dropped from the focus on the directrix. The 
conic may be of any of the following types : 

ellipse (e < 1), 
parabola (e = 1), 
hyperbola (e > 1). 

In the case of the hyperbola, (6.509) gives only the branch adja- 
cent to the focus. * 

Comparing (6.508) and (6.509), we note that it is always 
possible to bring the equations into complete agreement by 
choosing for I and e the values 



SBC. 6.5) MOTION OF A PARTICLE 179 



(6.510) I = , e - J 

/* V 



1 + 



2M 2 



Accordingly, we may say: The orbit described by a particle, 
attracted to a fixed center by a force varying as the inverse square 
of the distance, is a conic having the center of force for focus. The 
semi-latus-rectum and the eccentricity are given by (6.510) in terms 
of the angular momentum and energy per unit mass. The orbit 
may be of the following types: 

ellipse (E < 0), 
parabola (E = 0), 
hyperbola (E > 0). 

The fact that orbits may be classified thus in terms of the total 
energy is remarkable. 

The most important orbits in astronomy (those of the planets) 
are ellipses. Recurring comets describe orbits which are elon- 
gated ellipses, approximating to parabolas. A body with a 
parabolic or hyperbolic orbit would pass out from the solar 
system, never to return. 

Constants of the elliptical orbit. 

Let us now confine our attention to the elliptical orbit. Since 
the orbits of the planets are of this type, a great wealth of 
technical detail has been developed about the elliptical orbit. 
We shall here give only a brief treatment. 

It is evident from (6.509) that the shape and size of an orbit 
(but not its orientation in space) are determined by the two 
constants I, e. These are related to the constants E, h by 
(6.510). Thus, of the various constants which appear in our 
equations, we are to regard /x (the intensity of the force center) 
as given once for all, whereas the constants I, e, E, h take differ- 
ent values for different orbits. On account of (6.510), only two 
of these constants are independent. We may use as an inde- 
pendent pair any two which prove convenient. 

Instead of using (I, e) as fundamental constants, it is better 
to use (a, e), where a is the semiaxis major of the orbit. Now, 

(6.511) I - - a(l - e 2 ), 

6 being the semiaxis minor. We shall refer to (a, e) as the 
geometrical constants of an orbit and (E, h) as its dynamical 



180 



PLANE MECHANICS 



[SBC. 6.5 



constants. The formulas of transformation from one set to 
the other are as follows: 



(6.512) 



a = 



E = - 



2E 



2a 



2Eh* 



There is a simple formula giving the speed q at any point 
of the orbit in terms of the radius vector. By the equation of 
energy, we have 

to* - J - B. 

Substituting for E from (6.512), we obtain 



(6.513) 



The periodic time. 

We now ask: How long does the particle take to describe the 
elliptical orbit? This time is called the periodic time (T). We 
seek an expression for r in terms of the fundamental constants. 

The periodic time cannot be obtained from (6.414), because 
the time has been eliminated from this equation. We refer 
p instead to (6.408), which gives for 

the areal velocity 

A =ifc. 

If F is the focus at which the cen- 
ter of force is situated, it follows at 
once that the particle describes an 
arc VP, starting from the vertex* 
V nearer to F, in a time 2A/h, 
where A is the area of the sector subtended at F by this arc 
(Fig. 80). Following the point P right round the orbit, we get for 
the periodic time 




Fia. 80.- 



The rate of increase of A 
is constant. 



(6.514) 



2A 



* The vertex V is called perihelion the point closest to the sun the 
other vertex being called aphelion the point away from the sun. 



SBC. 6.5] MOTION OF A PARTICLE 181 

where A is now the total area of the ellipse. We might sub- 
stitute A = 7ra6; but, to be systematic, we should express r 
in terms of either the geometrical constants or the dynamical 
constants. Since 

6 = \/l e 2 , 
we obtain, by some easy calculations, 

<' ' - 



(We remember that E < for the elliptical orbit.) 

It is remarkable that the formula involves only one geometrical 
constant or one dynamical constant. All orbits with the same 
semiaxis major have the same periodic time; so also have all 
orbits with the same total energy. 

Kepler's laws. 

Before the mathematical theory given above had been devel- 
oped by Newton, Kepler deduced the following laws of planetary 
motion from a careful study of the results of astronomical 
observations: 

I. Each planet describes an ellipse with the sun in one focus. 

II. The radius vector drawn from the sun to a planet sweeps 
out equal areas in equal times. 

III. The squares of the periodic times of the planets are 
proportional to the cubes of the semiaxes major of their orbits. 

Starting from Newton's law of gravitation, we have shown 
that all these statements are true. But it is interesting to adopt 
the historical point of view and face the problem as it presented 
itself to Newton: Given Kepler's laws as a statement of fact, what 
is the law of gravitational attraction? 

Law II tells us that h the angular momentum per unit mass 
is constant, and hence that the force must be directed toward 
the sun. From Law I, we know that the equation of an orbit 
may be written 

u = -T (1 + e cos 0). 
Then by (6.414) the force per unit mass is 

(MID " 



182 PLANE MECHANICS [SEC. 6.5 

Thus for each planet the force varies inversely as the square of 
the distance. But it remains to prove that the force is of the 
form 

(6.517) mP 2, 

where m is the mass of the planet and /* a constant, the same for 
all the planets. To show this, we appeal to Law III. We 
know that for an elliptical orbit, described under a central force 
directed to a focus, 

2irab 

r = -/T' 
and so 



a 3 h*a h* ' 

But, by Law III, this is a constant, the same for all planets. 
Thus h*/l is the same for all planets; and so, by (6.516), P = /m 2 , 
where /* is the same for all planets. Hence (6.517) is true, and 
Newton's law of gravitation is thus deduced as a consequence 
of Kepler's laws. 

If Kepler's laws were accurately true, we should have to 
regard the sun as fixed and the planets as attracted only by 
the sun. More precise measurements show that Kepler's laws 
are only an approximation and that the inverse-square law of 
attraction holds for every pair of bodies. It is fortunate that 
the observations of Kepler's time were crude, because otherwise 
the simplicity of the law of gravitation would have been obscured. 

The two-body problem. 

An accurate dynamical treatment of the solar system involves 
complexities far greater than those encountered in the above dis- 
cussion. First, the sun is accelerated by the attractions of the 
planets; secondly, the mutual attractions of the planets influenqe 
their motions. A full treatment of the problem belongs to the 
subject of celestial mechanics, and we make no attempt to discuss 
it here. 

We may however ask: What is the behavior of two bodies which 
attract one another according to the law of the inverse square? This 



SEC. 6.5] 



MOTION OF A PARTICLE 



183 



problem presents itself in nature in the case of a double star and 
in the problem of the moon's motion relative to the earth, the 
attraction of the sun being neglected. 

We showed in Sec. 5.2 that, if there are no external forces, 
the mass center of a system moves in a straight lino with con- 
stant velocity, relative to a Newton- 
ian frame of reference. We can then 
take another Newtonian frame in 
which the mass center C is at rest. 
We suppose this done for the two-body 
problem in Fig. 81. 

Let m, m' be the masses of the par- 
ticles, r, r' their position vectors rela- 
tive to C, and i a unit vector drawn parallel to the line joining m r 
to m. Then the equations of motion are, in vector form, 




FIG. 81. The two-body 
problem. 



(6.518) 



mi = - 
m'r" 



Gmm' 



Gmm' . 



Now, from the definition of mass center, 

/ m + m' 



m 
7 



(6.519) 

hence (6.518) may be written 
GmM' 



r + r" 



m' 



(6.520) 



(m + m')*' 



f m'r' 



Gm'M . 

' !<> ^ 



M 



m 3 



(m + 



But these equations have the form of equations of motion under 
central forces varying as the inverse square of the distance. 
Therefore, each body moves about the fixed mass center as if attracted 
to it by the gravitational force due to a mass M' in the first case 
and a mass M in the second case. 

To find the motion of m relative to m', we note that the relative 
position vector is 
(6.521) R - r - r'. 



184 PLANE MECHANICS [SBC. 6.6 

Thus, by (6.518), 

(6.522) mR = m* - S n>V = - Gm(m R + m/) i, 

since R = r + r'. 

This result may be expressed as follows: TT&e motion of one of 
the bodies (m) relative to the other (m f ) takes place precisely as if 
the latter were fixed and its mass increased from m' to m + m f . 

It is evident from symmetry that, when a particle is attracted 
by a fixed center 0, its orbit lies in a plane, i.e., the plane contain- 
ing and the initial velocity vector. In the case of the two-body 
problem, the orbits both lie in one plane when viewed in a frame 
of reference in which the mass center C is fixed. But in any other 
Newtonian frame the motion appears very complicated. 

6.6. SUMMARY OF APPLICATIONS IN PLANE DYNAMICS MOTION 
OF A PARTICLE 

I. Ballistics. 

(a) No resistance; the trajectory is a parabola. 

(b) Resistance independent of height [R = mg4>(q)]] the 
trajectory may be found by quadratures when the following 
differential equation of the hodograph has been integrated: 



(6.601) or 

v ' cos 



(c) Resistance proportional to # 2 ; (6.601) may be integrated 
in terms of elementary functions, but the complete determination 
of the trajectory is very complicated. Motion in a vertical 
line is easily determined. 

(d) Resistance proportional to g; the trajectory is easily 
found. 

II. Harmonic oscillators. 

(a) Simple harmonic oscillations: 

(6.602) x + p*x = 0, 

(6.603) x = A cos pt + B sin pt, or x = a cos (pt + ) ; 

(T = = 2ir <J- for simple pendulum). 
r if v / 

(b) Oscillations with disturbing force mX: 

(6.604) X = constant; center of oscillation displaced. 



SBC. 6.6] MOTION OF A PARTICLE 185 

k 
(6.605) X = k cos ct\ x - a cos (pt + c) + 2 _ 2 cos ct. 



(c) Oscillations with damping ( 
(i) Light damping (/* < p) or oscillatory: 



(6.606) x = ae-*' cos (It + e), 

(Ratio of successive swings to opposite sides = e~ rft/l .) 
(ii) Heavy damping (/x > p) or deadbeat: 

(6.607) x = Ae-^-w + Be~^ +l)i y I = vV - P 2 . 

(d) Forced oscillations (periodic disturbing force mk cos ct) : 
after a long time the motion approximates to 

(6.608) x = 6 cos (ct + 17), 

where & and 17 are independent of the initial conditions. 

III. General motion under a central force (mP toward center). 
(a) Equations of motion: 

(6.609) * = - ^ y = - ^; 
d*u P 



(6.610) <d6* ' 

( h = r 2 6 = pg = twice the areal velocity ; 

(6.611) ( -JT ) + u 2 - TO ; (^ ~ constant total energy). 
\cLB / h u 

(b) Orbit symmetric with respect to apse line. 

(c) If P = k 2 r, the orbit is a central ellipse. 

IV. Planetary orbits. 

(a) The orbit under an attraction mn/r 2 is a conic section with 
one focus at the center of force: ellipse for E < 0, parabola for 
E = 0, hyperbola for E > 0. 

(b) For elliptical orbit 

(6.612) - = 1 + e cos 



(6.613) 



periodic time = r = 2?r ^/ 



186 PLANE MECHANICS [Ex. VI 

V. Two-body problem. 

Relative motion is the same as if one body were held fixed and 
its mass increased to the sum of the two masses. 

EXERCISES VI 

1. A particle is projected upward in a direction inclined at 60 to the 
horizontal. Show that its velocity when at its greatest height is half its 
initial velocity. (Neglect the resistance of the air.) 

2. A particle of mass ra moves on a straight line under the influence of a 
force directed toward the origin on the line and proportional to the dis- 
tance from 0] the force at unit distance is of magnitude wfc 2 . The particle 
passes O with a velocity M. If x is its coordinate at time t and v its velocity 
at that instant, show that r 2 + fc 2 s 2 - u*. 

3. Prove by a general argument, not involving any particular law of 
resistance, that a body thrown vertically upward in a resisting medium will 
return to the point of projection with a velocity less than that with which it 
was projected. 

4. A gun is mounted on a hill of height h above a level plain. Show 
that, if the resistance of the air is neglected, the greatest horizontal range for 
given muzzle velocity V is obtained by firing at an angle of elevation such 
that 

coscc 2 - 2(1 + gh/V 2 ). 

6. Find the greatest distance that a stone can be thrown inside a hori- 
zontal tunnel 10 feet high with a velocity of projection of 80 feet per second. 
Find also the corresponding time of flight. 

6. In a resisting medium two identical bodies are let fall from the same 
position at instants separated by an interval t. Show that the distance 
between them tends to the limit vt, where v is the limiting velocity. 

7. Calculate the rate of loss of energy (kinetic + potential) for a damped 
harmonic oscillator vibrating as in (6.330). 

8. A spring with compression modulus X supports a mass m. Show that 
the period of vertical oscillations under gravity is 2ir \/ml/\, where I is the 
natural length of the spring. (The compression modulus is the ratio* of the 
force producing compression to the compression per unit length.) 

9. A particle moves in a plane, attracted to a fixed center by a force 
varying as the inverse cube of the distance. Find the equation of the orbit, 
distinguishing the three different cases which may arise. 

10. A particle is attracted toward a fixed center by a force /i/r 2 per unit 
mass, M being a constant and r the distance from the center. It is projected 
from a position P with a velocity of magnitude q , making an angle a with 
OP. Assuming that OP < 2///0J, show that the orbit is an ellipse; determine 
(in terms of AC, q Q) a. and the distance OP) the eccentricity of the orbit and 
the inclination of the major axis to OP. 

11. A particle moves under the influence of a center which attracts with a 
force [(6/r 2 ) + (c/r 4 )], b and c being positive constants and r the distance 
from the center. The particle moves in a circular orbit of radius a. Prove 
that the motion is stable if, and only if, a*b > c. 



Ex. VI] MOTION OF A PARTICLE 187 

12. A particle of mass m moves in a central field of attractive force of 
which the intensity is 



where k is a constant. Prove that a circular orbit of radius r is stable if, and 
only if, r 2 < \. 

13. Deduce the following relations for an elliptical orbit under the New- 
tonian law of attraction: 

r a(l e cos E), 
M E - e sin E, 

where r is the radius vector drawn from the center of attraction, a the 
semiaxis major of the orbit, E the eccentric anomaly, and M the mean 
anomaly. (These anomalies are angles, defined as follows. Let O be the 
geometrical center of the orbit, V its perihelion, and P the position of the 
particle at time t. Then E is the eccentric angle of P vanishing when P 
is at V. To define M, we consider a point moving on the orbit with con- 
stant angular velocity about 0, starting from V with P and completing the 
circuit in the actual periodic time. If Q is the position of this point at time 
t, then the value of Af corresponding to P is the angle QOF.) 

14. A simple pendulum of mass m and length a is hanging in equilibrium. 
At time t = a small horizontal disturbing force X comes into operation and 
continues to act, varying with time according to the formula 

X mb sin 2pt, 

where p 2 g/a. Find a formula giving the position of the pendulum at any 
time. 

15. A body of mass m is projected vertically upward in a medium for 
which the resistance is mk*v*. If the initial velocity is t> , show that the 
body returns to the point of projection with a velocity v\ such that 



. 
1 g + k*vl 

16. Mud is thrown off from the tire of a wheel (radius a) of a car traveling 
at a speed F, where V 2 > ga. Neglecting the resistance of the air, show 
that no mud can rise higher than a height 

, V* , ga* 
+25 + 2F* 

above the ground. 

17. A shell is fired vertically upward with speed 90. The resistance is 
mgCq*. Show that it attains its greatest height at time I, given by 

tan (gt v/C) - flo VU. 

Deduce that, no matter how large go may be, t cannot exceed ^""'C""*. 

18. A particle of mass m describes an elliptical orbit of semiaxis major a 
under a force mp/r* directed to a focus. Prove that the time average of 



188 PLANE MECHANICS [Ex. VI 

reciprocal distance is 



1 rdt I 
- i =5 - 

r J r a 



and deduce that the time average of the square of the speed is 



The integrals are evaluated for a complete revolution. 

19. A bomb is dropped from an airplane flying horizontally at a height h 
with speed U. Assuming the linear law of resistance R = mgCq as in 
(6.225), and further assuming that this resistance is small, show that the 
time of fall is approximately 



Show also that the horizontal distance through which the bomb falls is 
approximately 



20. In the problem of two bodios attracting according to the inverse square 
law, there are four orbits: the orbits of cither body relative to the other and 
the orbits of either body relative to the mass center. Show that all four 
orbits have simultaneous apsides and the same eccentricity. 

21. Two particles of masses m, m' distant a apart are projected with 
velocities q, q', respectively; the directions of projection and the line joining 
the particles are mutually perpendicular. Find the condition that the 
relative orbits under their mutual attraction may be ellipses; assuming the 
condition to be satisfied, find the periodic time. 

22. The motion of an oscillator may be represented graphically in a plane, 
x being shown as abscissa and x as ordinate. The history of the oscillator 
is then a curve. Show that for an undamped harmonic oscillator this 
curve is an ellipse, and for a lightly damped oscillator it is a curve spiraling 
in to the origin. Investigate the curve for a heavily damped oscillator, and 
show that it will be a straight line through the origin for special initial 
conditions. 



CHAPTER VII 

APPLICATIONS IN PLANE DYNAMICS- MOTION OF A 
RIGID BODY AND OF A SYSTEM 

7.1. MOMENTS OF INERTIA. KINETIC ENERGY 
AND ANGULAR MOMENTUM 

Definition of moment of inertia and some direct calculations. 

The moment of inertia of a particle about a line is defined 
as / = mr 2 , where m is the mass of the particle and r its per- 
pendicular distance from the line. The moment of inertia of a 
system of particles is defined as the sum of the moments of 
inertia of the separate particles. Thus, 

(7.101) / = 2} m ' r *<> 

i-i 

if the system consists of n particles of masses mi, m 2 , m nt 
situated at distances ri, r 2 , r n from the line about which the 
moment of inertia is taken. 

It is evident that the method of decomposition is applicable 
to moments of inertia. Thus, if a system is split into two parts 
with moments of inertia /i and /2, the moment of inertia of the 
complete system is 

(7.102) / = /i + Ii. 

It is convenient to define a length k called the radius of gyration. 
If a system of total mass m has a moment of inertia /, the radius 
of gyration k is defined by the equation 

(7.103) mk* = /. 

When the system consists of a single particle, it is evident that 
the radius of gyration about any line is simply the distance from 
the line. 

In the case of a continuous distribution of matter, the definition 
(7.101) passes over into 

(7.104) / = JY 2 dm, 

where the integration sign indicates the limit of a process in 
which the system is divided into a great number of very small 
parts, and the sum taken; dm is the mass of an infinitesimal 

189 



190 PLANE MECHANICS [SEC. 7.1 

element, and r is its distance from the line about which the 
moment of inertia is to be found. 

Moment of inertia has the dimensions [ML*] and is measured in gm. cm. a 
in the c.g.s. system and in Ib. ft. 2 in the f.p.s. system. The square of the 
radius of gyration has the dimensions [moment of inertia]/[mass], or [ML*]/ 
[M], i.e., [L 2 ]. Thus radius of gyration has dimensions [L] and so is a 
length. 

Let us now calculate some simple moments of inertia. 

Hoop. It is evident that the moment of inertia of a thin 
hoop of mass m and radius a about a line through its center 
perpendicular to its plane is ma 2 . 

Rod. Let us calciilate the moment of inertia of a uniform rod 
of mass m and length 2a about a line through its center per- 
pendicular to its length. Taking the rod for or-axis, with the 
origin at the center of the rod, the mass of an element dx is 

, m dx 
dm = ^ 
2a 

Hence, by (7.104), we have 

m dx ., 



(7.105) 



2a 



Rectangular plate. Consider now a uniform rectangular plate 
of mass m and edges of lengths 2a, 2b. We wish to calculate 
the moment of inertia about the line in its plane passing through 
the center and parallel to the edge 26. We imagine the plate 
split into thin strips parallel to the edge 2a. Let dm be the 
mass of a strip. Then the moment of inertia of the strip is 
Jo 2 dm, by (7.105), and hence the moment of inertia of the whole 
plate is -ywa 2 . 

Circular disk. We wish to find the moment of inertia of a 
uniform circular disk of mass m and radius a about a line through 
its center perpendicular to its plane. We imagine the disk split 
up into thin rings by a great number of circles concentric with the 
boundary. If r, r + dr are the inner and outer radii of a ring, 
the area of the ring is 2irr dr, and its mass is 

, 2irr dr 

dm = m 5 

ira 2 

The moment of inertia is 

(7.106) / = r 2 dm = r 9 dr 



SEC. 7.1] MOTION OF A RIGID BODY AND OF A SYSTEM 191 



Circular cylinder. The moment of inertia of a solid circular 
cylinder about its axis follows immediately from (7.106). For 
we may imagine the cylinder split up by planes perpendicular 
to its axis into a great number of thin disks. When we add 
together their moments of inertia, we 
get 7 = iwa 2 , where m is the mass of 
the cylinder and a its radius. The 
length of the cylinder does not appear 
explicitly in the formula. 

Sphere. To find the moment of in- 
ertia of a solid sphere of mass m and 
radius a about a diameter, we imagine 
it split into thin circular disks by 
planes perpendicular to the diameter 
in question. Figure 82 shows the sec- 
tion of the sphere by a plane through 
the diameter (Ox) about which the moment is to be calculated. 
If p is the density of the material, the mass 'of the disk between 
planes at distances #, x + dx from the center is 

p iry 2 dx, 

where y is the radius of the disk. By (7.10G), the moment of 
inertia of the disk is 

dl = ^Trpy 4 dx. 
But y 2 = a 2 # 2 , and so 




FIG. 82. Solid sphere split 
into thin circular disks for the 
calculation of moment of 
inertia. 



But 



= ITTP f a ( - X 2 
J a 



m = 



and so the moment of inertia of the sphere is 
(7.107) I = |ma 2 . 

Exercise. Show by the theory of dimensions, without calculations, that 
the above moments of inertia of the hoop, the rod, the circular disk, the 
circular cylinder, and the sphere are all necessarily of the form Cma z , where 
C is a pure number. 

Theorem of parallel axes. 

The theorem of parallel axes gives us an easy method of cal- 
culating the moment of inertia of a system about any line, 
when the moment of inertia about a parallel line through the 
mass center is known. Figure 83 shows a projection onto a 



192 



PLANE MECHANICS 



[SEC. 7.1 



plane perpendicular to the two lines, 0' being the projection of 
the line through the mass center and the projection of the 

other line. Introducing parallel co- 
ordinate axes as shown, let (a, 6) be 
the coordinates of 0' relative to 0. 
If x, y are the coordinates of any 
point relative to the axes Oxy, and 
, x', y' the coordinates of the same 



O 



a 



FIG. 83. Coordinates for tho 
proof of the theorem of parallel 
axes. 



point relative to O'x'y', then 
(7.108) 



a, 
b. 



With the notation used at the 
beginning of this section, the moments of inertia about the lines 
through and 0' are, respectively, 



(7 109) 



m t (z? + </?), 7' = 



i- 



+ yft. 



Then, by (7.108), 



(7.110) 



I' + m(a z 



where m is the total mass of the system, since 

Li n 




-2a 



FIG. 84. Rod and sphere: 
the moment of inertia about L 
is required. 



from the definition of mass center. 
We may state (7.110) in words as 
follows: The moment of inertia of a 
system about an axis L is equal to the 
moment of inertia of the same system 
about an axis through the mass center 
parallel to L, together with the moment 
of inertia about L of a particle with a 
mass equal to the total mass of the sys- 
tem, placed at its mass center. 



As an application of this theorem of parallel axes, we note that 
by (7.105) the moment of inertia of a rod of length 2a about a 
line through one end, perpendicular to the rod, is 



SBC. 7.1] MOTION OF A RIGID BODY AND OF A SYSTEM 193 

As another application, suppose we wish to find the moment of 
inertia about the line L of the apparatus shown in Fig. 84, 
consisting of a rod of mass m and length 2a, with a sphere of mass 
M and radius b attached to the end of the rod. From (7.105) and 
(7.107), combined with the theorem of parallel axes, we obtain 
at once 



(7.111) / = ma 2 + M[ffe 2 + (2a + fc) 2 ]. 

Theorem of perpendicular axes. 

The theorem of perpendicular axes is useful for the calculation 
of moments of inertia of plane distributions of matter. Let 
Oxyz be rectangular axes, and let there be a distribution of 
matter in the plane = 0. Denoting by A, B, C the moments 
of inertia about the three axes, we have 

(7.112) A 

Obviously, 

(7.113) C - A + B; 

this result constitutes the theorem of perpendicular axes. 

As an application, suppose we wish to find the moment of 
inertia of a rectangular plate of edges 2a, 26, about a line through 
its center perpendicular to its plane. Taking the origin at the 
center, the axes Oxy parallel to the edges, and the axis Oz per- 
pendicular to the plate, we have, as already established, 



(7.114) A = %mb\ B = 
Hence the required moment of inertia is 

(7.115) C = A + B = m(a 2 + 6 2 ). 

More information about moments of inertia will be found in 
Sec. 11.3. 

Exercise. The moment of inertia of a hoop of mass m and radius a 
about a diameter is $nia 2 and that of a circular disk of the same mass 
and radius about a diameter is imo 2 . Verify these statements. 

Kinetic energy and angular momentum. 

As we have seen in Sec. 5.2, kinetic energy and angular momen- 
tum play an important part in the dynamics of systems. We 



194 PLANE MECHANICS [SBC. 7.1 

shall now show how these quantities are to be calculated when 
the system is a rigid body moving parallel to a plane. 

Let us first suppose that the rigid body is rotating about a 
fixed axis. Let o> be the instantaneous value of the angular 
velocity. Then the kinetic energy of a particle of mass m v 
situated at a distance r- from the axis is raT 2 a> 2 , and so the kinetic 
energy of the rigid body (supposed to consist of n particles) is 

(7.116) T = |a 

where I is the moment of inertia about the axis. The angular 
momentum of a particle about the axis of rotation is w t r 2 o>, and 
so the angular momentum of the rigid body is 

(7.117) h = < 

Let us now suppose that the rigid body no longer rotates about 
a fixed axis but moves in a general manner parallel to a fixed 
fundamental plane (cf. Sec. 4.2). Let us imagine an observer 
traveling with one of the particles (A) of the rigid body and 
observing the motion of the particles relative to him. He can 
compute a relative kinetic energy and a relative angular momen- 
tum about a line through A perpendicular to the fundamental 
plane, using in these computations the velocities of the particles 
relative to him. Since relative to him the body rotates with ang- 
ular velocity o> about a fixed axis through A, the formal calcu- 
lations are precisely as above and lead to formulas (7.116) and 

(7.117) for the relative kinetic energy and angular momentum. 
Although this is true for any particle A of the body, the results 

are most useful when A is the mass center. Let us restate them: 
The kinetic energy and angular momentum, both relative to the 
mass center, of a rigid body moving parallel to a plane are 

(7.118) T = i/co 2 , h = Jw, 

where co is the angular velocity of the body and I its moment of 
inertia about an axis through the mass center perpendicular to the 
plane of motion. 

The following theorem of Konig enables us to complete the 
calculation of the kinetic energy of a rigid body moving parallel 
to a plane. We shall prove it in general three-dimensional form. 



SEC. 7.1] MOTION OF A RIGID BODY AND OF A SYSTEM 195 

THEOREM OF KONIG. The kinetic energy of a moving system is 
equal to the sum of (i) the kinetic energy of a fictitious particle 
moving with the mass center and having a mass equal to the total 
mass of the system and (ii) the kinetic energy of the motion relative 
to the mass center. 

Let Oxyz be fixed axes and O'x'y'z' parallel axes through the 
mass center. Let x, y, z be the coordinates of the mass center 
referred to Oxyz. Then, for any particle, we have 

(7.119) x t = x + x't, . y t = y + y' i} z, = z + zj. 
The kinetic energy of the system is 



(7.120) T = 



Let us differentiate (7.119) and substitute for z, y if zt in (7.120). 
Certain terms vanish on account of the relations 



(7.121) Y m& = 2) mtf % = 

ff\ ff\ i 



which are consequences of the definition of mass center given in 
Sec. 3.1. We obtain 

(7.122) T = im(x 2 + # 2 + I 2 ) + ^ i) m t (x? + y? + z?), 



where m = V m t , the total mass of the system. Thus the 



theorem is proved. 

It is convenient to refer to the kinetic energy of the fictitious 
particle as the "kinetic energy of the mass center," so that 
our result reads 

(7.123) T = T + T', 

where T Q is the kinetic energy of the mass center and T' the 
kinetic energy relative to the mass center. This general result 
holds even though the system is not a rigid body. 
In the case of a rigid body, we have the important formula 

(7.124) T 



196 PLANE MECHANICS [Sic. 7.2 

where m = mass of body, 

q = speed of mass center, 

/ = moment of inertia about mass center,* 

co = angular velocity. 

Exercise. Find the kinetic energy of a disk of mass m and radius o, 
rolling along the ground with speed q. 

7.2. RIGID BODY ROTATING ABOUT A FIXED AXIS 
General methods. 

In Sec. 5.2 we developed the principle of angular momentum 
(5.214) and the principle of energy (5.223). Let us insert in these 
equations the values of h and T given in (7.117) and (7.116); 
then we have 

(7.201) /co = N (principle of angular momentum), 

(7.202) l/co 2 + V = E (principle of energy). 

These equations represent the two fundamental methods of 
finding the motion of a rigid body which turns about a fixed 
axis. Let us recall the meanings of the terms: 

/ = moment of inertia about the fixed axis, 

w = angular velocity, 

N = moment of external forces about the fixed axis, or torque. 

V = potential energy, 

E = total energy (a constant). 

It must be remembered that (7.201) is always valid; (7.202), 
on the other hand, holds only when the system is conservative. 
It would not hold, for example, if there were a frictional torque. 

Flywheels. 

Let us consider a flywheel rotating about a fixed axis which 
passes through its mass center. Gravity contributes nothing to 
the moment about the axis and so does not influence the motion. 
We suppose the torque N supplied by a motor or brakes. Since 
the forces involved here will not, in general, be conservative, we 
use (7.201) as the equation of motion; so we write 

(7.203) Jco = N. 

We may note the resemblance between this equation and the 

* More precisely, the moment of inertia about the line through the mass 
center perpendicular to the plane of motion. 



SEC. 7.2] MOTION OF A RIGID BODY AND OF A SYSTEM 197 

equation of motion of a particle moving on a straight line, 

mu = X; 

mass corresponds to moment of inertia, linear velocity to angular 
velocity, force to torque. This mathematical similarity may be 
used to solve a problem in the dynamics of a rotating flywheel, 
when the solution of the analogous problem for a particle moving 
on a straight line is already known. 

If the torque N is constant, (7.203) gives 

(7.204) co = y t + A, 

where A is a constant of integration; hence, if is the angle 
turned through, we have 6 = co and 

(7.205) = ^ * 2 + At + B, 

where B is another constant of integration. This motion is 
analogous to the motion of a particle under a constant force. 

The importance of the flywheel in machinery lies in its capacity 
to smooth out motion. In a steam or gasoline engine the torque 
is not uniform, and without a flywheel (or something equivalent) 
the motion would be jerky. As an illustration, let us work out 
the case where a flywheel of moment of inertia 7 is under the 
action of a torque with a fluctuating part, 

N = N Q + Ni cos ct, (No, Ni constants), 

and a load proportional to angular velocity. The equation of 
motion is 

(7.206) 7w = No + Ni cos ct - Lo>, 
the last term corresponding to the load; thus 

(7.207) w + -ju = ~ (No + Ni cos ct). 

This is a standard type of differential equation, which has the 
integrating factor e u/I ; the solution is 

(7.208) o) = e~ Lt/I I A + j f e Lt "(N Q + Ni cos ct) dt\, 

where A is a constant of integration. Using R to denote "real 
part of," we have 



198 



PLANE MECHANICS 



[Sec. 7.2 



(7.209) J* e w " cos ctdt = RJ* 



_. e (L//+tc) 



= 72 

= e Lt R ^7^72 (cos * + * sin ct} 
_ c n// cos (ct + ) 

((L/iy + c']*' 
where is a constant; its value is of no present interest. Hence, 



(7.210) = Ac + + 



c , ]t 



cos (ct + .). 



The first term dies away as t increases. The ratio of the ampli- 
tude of the third term to the second term is 

(7.211) ~ 

By increasing the moment of inertia of the flywheel, we can make 
this ratio as small as we please and so approximate to the steady 
motion co = N /L, even though the fluctu- 
ating part Ni cos ct may be greater in mag- 
nitude than the steady torque No. 

The compound pendulum. 

In Sec. 6.3, we discussed the small oscil- 
lations of a simple pendulum, consisting of 
a heavy particle attached to a fixed point by 
a light string. We shall now discuss the 
compound pendulum, which is a rigid body 
free to oscillate under the influence of grav- 
ity about a fixed horizontal axis. We shall 
use the equation of energy (7.202), but the 
results may be obtained with equal ease 
from (7.201). 

In Fig. 85 the plane of the paper is the plane through the 
mass center C perpendicular to the axis of suspension. The axis 
cuts it at 0, which is called the point of suspension. We shall 
use the following notation: 

a = OC, 

m = mass of pendulum, 

k e = radius of gyration about (7, 

fco = radius of gyration about 0. 




Fio. 85. A compound 
pendulum. 



SEC. 7.2] MOTION OF A RIGID BODY AND OF A SYSTEM 199 

If 6 denotes the inclination of OC to the vertical, the potential 
energy is 

V = mga cos 0, 
and (7.202) gives 

(7.212) $mk* Q 6* - mga cos 6 = E, 

where E is a constant. 

This equation gives the angular velocity at any position, 
when E has been found from the initial conditions. 

For example, if the pendulum starts with C directly below 
and with angular velocity w , we have 

E = Tfmkfa* mga] 
equation (7.212) gives 

ffy O1 O\ ^i2 2 "J/** 1 * 91/1 

(7.213) 2 = cog -TJ- sin 2 TfO. 

This will vanish when takes the values a, where 



and so the pendulum oscillates through the range ( ,). 

Since sin 2 -^0 cannot exceed unity, it is evident from (7.213) 
that 6 never vanishes if 



if started with such an angular velocity, the pendulum travels 
right around. 

Differentiation of (7.212) gives 

(7.214) klS + ga sin = 0, 

as an alternative form for the equation of motion of a compound 
pendulum. Had we used (7.201), we should have obtained this 
equation directly without differentiation. 

For small oscillations, we replace sin 6 by $ and obtain the 
solution 

(7.215) = a cos (pt + e), 

where a, are constants of integration and p 2 * ga/k\. This 



200 PLANE MECHANICS [SEC. 7.2 

is a simple harmonic motion with periodic time 

(7.216) T = ?* = 

P 

The simple pendulum is a special case of the compound 
pendulum; for a simple pendulum of length Z, we have ko = Z, 
a = Z, and so (7.214) gives 

(7.217) 16 + g sin 6 = 0, 

as the general equation of motion of a simple pendulum. The 
equation (6.304) was valid only for small oscillations. 

If we compare the motion of a compound pendulum, given by 
(7.214), with the motion of a simple pendulum, given by (7.217), 
we note that the two equations are mathematically identical 
provided that 

(7.218) I = ^- 

Thus, corresponding to any compound pendulum, we can con- 
struct a simple pendulum of length given by this formula, which 
will oscillate in unison with the compound pendulum; it is called 
the equivalent simple pendulum. 

Exercise. Show that a square plate of side 26 suspended from one corner 
oscillates in unison with a simple pendulum of length (4 \/2/3)6 = 1.896. 

Let us now suppose that a rigid body is given, with a number 
of thin parallel holes drilled through it. We can form a com- 
pound pendulum by passing an axis of suspension through any 
one of the holes. How does the periodic time of small oscillations 
depend on the position of the hole chosen? 

To answer this question, we note that, by the theorem of 
parallel axes (7.110), 

(7.219) kl - a 2 + /c?. 

Hence the formula (7.216) for the periodic time may be written 

(7.220) r 2 = - 

If we change the position of the point of suspension in the body, 
a changes but k c does not change. We see that T tends to 



SBC. 7.3] MOTION OF A RIGID BODY AND OF A SYSTEM 201 

infinity if a tends to zero or if a tends to infinity. Thus, we 
can obtain very slow oscillations by moving the point of sus- 
pension close to the mass center or far from it. 

On differentiating (7.220) with respect to a, we get 

d f 2 , 47T 2 A *J\ 

j~ ( T ) == - 11 -- 1 r 
da v ' g \ a 2 / 

Thus the periodic time is a minimum when the point of suspension 
is at a distance from the mass cen- 
ter equal to the radius of gyration 
about the mass center. 



We now ask whether it is pOS- FIG. 86. The periodic time is the 



sible to shift the point of suspen- m ' to U8pd " 8 



sion from a position to a new 
position O f (Fig. 86) on the line OC without changing the periodic 
time. With OC = a, CO' = b, the condition for equality of 
periodic times is, by (7.220), 

fr 2 If 2 

I W/ C J, L C 

+ ^ = b + r 

which is satisfied if 

(7.221) ab = fc 2 - 

The point 0', related in this way to the point of suspension 0, 
is called the center of oscillation. 

7.3. GENERAL MOTION OF A RIGID BODY PARALLEL 
TO A FIXED PLANE 

General methods. 

Probably the most useful principle available for the solution 
of problems in mechanics is the principle of energy in the form 
(5.223), namely 

(7.301) T + V = E. 

One must of course make sure, before attempting to apply 
this principle, that the system is conservative, i.e., that it has 
a potential energy V. 

When the system consists of a single rigid body moving parallel 
to a fixed plane, we may write (7.301) in the form 

(7.302) im<Z 2 + $Io> 2 + V = E, 



202 PLANE MECHANICS [Sac. 7.3 

where m = mass of body, 

q = speed of mass center, 

/ = moment of inertia about mass center, 

a) = angular velocity of body. 

However, (7.302) is only one equation. Sometimes we require 
more equations, and then we may employ the principles of 
linear and angular momentum in the forms (5.209) and (5u219). 
If Oxy are fixed axes in the fundamental plane, we have 

(7.303) mx = X, my = 7, /ci = N, 

where x, y are the coordinates of the mass center, X, Y are the 
total components of external forces in the directions of the axes, 
and N is the total moment of external forces about the mass 

center. Of the four equations 
(7.302) and (7.303), at most three 
are independent. 

Cylinder rolling down an inclined 
plane. 

Consider a cylinder of mass m 
and radius o, rolling down a plane 
inclined at an angle a to the hori- 
FIG. 87 -^^^ down an zontal (Fig. 87). We wish to de- 
termine the motion, and as an il- 
lustration we shall do so by two methods, first using the principle 
of energy and then the principles of linear and angular momen- 
tum. We assume the mass center to be situated on the geo- 
metrical axis of the cylinder. 

Let x be the displacement at time t of the center of the cylinder 
from its initial position at rest at t = 0, and 6 the angle through 
which it has turned. Then, by the condition of rolling, 

(7.304) x = aO. 

If A; is the radius of gyration of the cylinder about its axis, its 
kinetic energy is 

(7.305) T = \m& + 
or, by (7.304), 

(7.306) 




SBC. 7.3] MOTION OF A RIGID BODY AND OF A SYSTEM 203 

The potential energy is 

(7.307) V = mgx sin a. 
Hence, by (7.301), 

(7.308) & 



2 ) x* mgx sin a 



where E is the constant total energy. Actually E = 0, since 
x = = for t = 0. Differentiating (7.308) with respect to t, 
we obtain 



(7.309) 



g sin a. 



Thus the cylinder rolls down the inclined plane with a constant 
acceleration. 

A particle would slide down a smooth plane of inclination a 
with an acceleration g sin a. The value given by (7.309) is 
always less than g sin #, except in the limiting case k = 0, 
which corresponds to a concentration of all the mass of the 
cylinder on its axis. If the cylinder 
is a thin shell, we have k = a, and 
hence 



(7.310) 



x = 



sn a. 




If the cylinder is solid and uniform, 
we have k 2 = a 2 , and hence 

(7.311) x = |0 sin a. 

FIG. 88. External forces acting 

The method of energy does not on c y hnder - 

tell us the reaction between the cylinder and the plane, or how 
rough the plane must be in order that slipping may be avoided. 
To find out these things, we turn to the principles of linear and 
angular momentum, using (7.303). 

The reaction of the plane on the cylinder may be resolved into 
a normal component N and a component F in the plane (Fig. 88). 
These forces, with the weight mg, form the complete system of 
external forces. Thus, we have 



(7.312) 



' mx = mg sin a F, 
= mg cos a N, 
[ mk*'6 = Fa, 



204 PLANE MECHANICS [Sue. 7.3 

the second equation coming from resolution perpendicular to the 
plane. Using (7.304) and eliminating F, we get 

g sin a 



as in (7.309). Hence the components of the reaction are 

/^o^\ r fc 2 . mgk 2 sin a ,, 

(7.314) F = m -5 x = % , i N = mg cos a. 

fl O ~t~ AC 

In order that rolling may occur without slipping, we must have 
F/N ^ M, or 

/^oie\ -^ fc 2 tan a 

(7.315) p > - , 



where /* is the coefficient of static friction as in (3.202). 

Self-propelled vehicle. 

Consider an automobile (Fig. 89). The external forces acting 
on it are 

(i) gravity, 

(ii) the reactions of the ground on the wheels, 

(iii) resistance of the air. 

Without knowing any further details, we can apply the principle 
of linear momentum in the form (5.209) to the complete auto- 





Fio. 89. Automobile. 

mobile. If its mass is m and it is traveling on a horizontal road 
with acceleration f , then mi equals the total horizontal component 
of ground reactions and air resistance. The total vertical 
component of gravity, ground reactions, and air resistance is 
zero. 

Application of the principle of angular momentum in the 
form (5.219) requires a little care. For simplicity, we shall 
suppose that the wheels have no mass and hence no angular 
momentum. The angular momentum of the automobile about 



SEC. 7.3] MOTION OF A RIGID BODY AND OF A SYSTEM 205 



its mass center is then zero. Hence the total moment about the 
mass center of ground reactions and air resistance is zero. 

We can use the above results to find the greatest possible 
acceleration of an automobile on a street for which the coefficient 
of friction between ground and tire is ju. Since, by hypothesis, 
the mass of each wheel is zero, the rate of change of angular 
momentum of a wheel about its center is zero; hence the total 
moment of external forces on a wheel is zero. These forces 
consist of a force exerted by the axle, a couple due to engine or 
brakes, and a ground reaction. If the couple is absent, the 
ground reaction can have no moment about the center of the 
wheel. In fact, in the case of a wheel without mass, undriven and 
unbraked, the ground reaction has no frictional component. 



No 




FIG. 90. External forces acting on automobile. 

Figure 90 shows the external forces acting on an automobile, 
driven through the rear wheels on the right, air resistance being 
neglected. 
Let h = height of mass center above ground, 

61 = distance of front axle in front of mass center, 

62 = distance of rear axle behind mass center, 

Ni resultant of vertical reactions at two front wheels, 
# 2 = resultant of vertical reactions at two rear wheels, 

F = resultant of frictional forces at two rear wheels. 
Then, 



(7.316) 



tntf = F, 

) = tfi + N 
( = b z Nz - 



- mg, 
iJfi - hF. 



Solving for Ni, Nz, F, we obtain 

(7W^ N. - m gb * "^ N 
V/.oi/; ivi m -r , T > M 



F = mf. 



By the law of static friction (3.202) F/N 2 ^ M, and so 



206 PLANE MECHANICS [Site. 7.3 



or 
(7.319) 



+ 6 2 



This fraction represents the greatest acceleration possible 
without slipping. 

The maximum negative acceleration obtainable by the applica- 
tion of brakes, without slipping between the tires and the ground, 
may be found in a similar way. 

Internal reactions. 

The principles of linear and angular momentum do not involve 
the internal reactions between the particles of a rigid body. 
Nevertheless these reactions exist, and when they become 
excessive the body may break. As we shall now see, the prin- 
ciples of linear and angular momentum may be used to find the 
reactions. D'Alembert's principle (cf . Sec. 5.2) may also be used. 





o o 

FIG. 91. (a) A rod rotating about one ond. (6) Reactions on the portion BA. 

The essential point to note is that, when internal reactions 
are sought, the dynamical system considered is only part of the 
rigid body. We shall illustrate the method with an example. 

Figure 91a shows a uniform rod OA rotating about O with 
angular velocity w, which we shall first suppose to be constant. 
B is any point in the rod. We seek the reaction across the section 
of the rod at B. As in Sec. 3.3, the reaction exerted by OB on 
BA consists of a tension T, a shearing force S, and a bending 
moment M (Fig. 916). Let us regard gravity as non-existent. 
Then T, S, M constitute the whole system of external forces 
acting on BA. 

Let OA = Z, OB r. The acceleration of the mass center 
of BA is of magnitude $(l + r)w 2 , directed along AB. If m 



SBC. 7.4] MOTION OF A RIGID BODY AND OF A SYSTEM 207 

is the mass of the rod, the mass of BA is m(l r)/L Hence the 
principle of linear momentum, applied to BA as a dynamical 
system, gives 

(7.320) T - * ^f- (I 2 - r), 5 = 0. 

The angular momentum of J5A about its mass center is constant. 
Hence, by the principle of angular momentum, M = 0. Thus, 
when w is constant, the reaction in the rod at B is a tension, as 
given by (7.320). As a check, we note that T = for r = I, 
and T = $mo)H for r = 0. 

Let us now consider the more general case where w is a variable 
function of t. By (4.107) the acceleration of the mass center of 
BA has components 

(7.321) %(l + r)w 2 along AB, %(l + r)ci perpendicular to AB. 
Hence, 

(7.322) T = * ^ (Z 2 - r 2 ), 8 = i ^ (J 2 - r 2 ). 

If fc is the radius of gyration of BA about its mass center, the 
angular momentum of BA about its mass center is mk*u(l r)/L 
Hence, by the principle of angular momentum, 



(7.323) M - iS(l _ r) - (I - r). 
Thus, 

(7.324) M = OTti>a z ~ r) [** + W - r*)]. 

But 

fc 2 = Ad - r) f , 
and so 

(7.325) M = i ?y (I - r)*(2J + r). 

Exercise. Find where the rod is most likely to break, assuming that this 
occurs where the bending moment is greatest. 

7.4. NORMAL MODES OF VIBRATION 

Degrees of freedom. 

The position of a simple pendulum is determined by the value 
of one variable, namely, its inclination to the vertical, or the 
horizontal component of the displacement of the bob. A system 



208 PLANE MECHANICS [SBC. 7.4 

whose position may be specified by one variable or coordinate 
is said to be a system with one degree of freedom. 

A rod which can move in a plane, with one end constrained to 
move on a fixed line, can be described as to position by two vari- 
ables, namely, the distance of the constrained end from a fixed 
point on the constraining line and the inclination of the rod to 
the line. Each of these variables can take arbitrary values. 
A system whose position may be specified by two arbitrary and 
independent variables or coordinates is said to be a system with 
two degrees of freedom. 

Similarly, there are systems with n degrees of freedom, where 
n = 3, 4, - - ; 

In Sec. 6.3 we discussed the oscillations or vibrations of a simple 
pendulum and in Sec. 7.2 those of a compound pendulum. Each 
of these is a system with one degree of freedom. We now proceed 
to discuss systems with two degrees of freedom. 

Particles on a stretched string. 

Let there be a light elastic string of length 3a, stretched 
between points A, B. Let two particles, each of mass m, be 





a a 

FIG. 92. Loaded string vibrating. 

attached to the string at the points of trisection. For simplicity, 
we shall neglect gravity; or, equivalently, we may suppose the 
particles supported on a smooth horizontal plane. 

Initially the particles are at rest and the tension in the string 
is a constant (S) throughout. The particles are given small dis- 
placements perpendicular to the string and then released. We 
wish to investigate the resulting oscillations. 

Figure 92 shows the situation at time t. The particles are 
at C and D, their displacements from the positions of equilibrium 
being denoted by x and y, which are small quantities. The 
inclinations of the portions of the string to AB are small, of the 
same order as x and y. Hence, since the cosine of a small 
angle differs from unity by a small quantity of the second order, 
it is seen that the lengths AC, CD, DB are each equal to a, to 



SBC. 7.4] MOTION OF A RIGID BODY AND OF A SYSTEM 209 

the first order of small quantities inclusive. Hence the tensions 
in these portions are equal to S to this order. 

Resolving forces in the direction perpendicular to AB, we 
obtain the following equations of motion: 

(7.401) mx = -2-s5JZJ>, m y = S 5-H2 - S 

a a a a 

Writing 

(7.402) k* = , 

' ma 

we simplify these equations to 

(7.403) * + 2k 2 x - k z y = 0, -k 2 x + y + 2k 2 y = 0. 
We try solutions of the form 

(7.404) x = A cos (nt + e), y = B cos (TI + *), 

where -4, #, n, e are constants. The equations (7.403) are 
satisfied provided A, B, n satisfy the equations 



(7-405) l_Ai* + *(- + 2fc)=0. 

Elimination of A and B gives the determinantal equation 

n 2 - 2k 2 k* 

k 2 n 2 - 2k 2 



(7.406) 



or 

(7.407) n 4 - 4fc 2 n 2 + 3fc 4 = 0; 
the solutions are n\, n 2 , where 

(7.408) n\ = A; 2 , n - 3A; 2 . 

When n is known, either of the equations (7.405) gives, for the 
ratio B/A, 

(7.409) I = 2 ~ F 



Thus for n = m, 5/-A = 1; and for n = n 2 , J5/A = 1. 

Hence if Ai, 1 are arbitrary constants, the following is a 
solution of (7.403) : 

(7.410) x = Ai cos (fa + i), y = Ai cos (fa + 1). 



210 



PLANE MECHANICS 



[SBC. 7.4 



Also, if A 2, 2 are arbitrary constants, the following is a solution 
of (7.403): 

(7.411) x = A* cos (kt V3 + * 2 ), 2/ =* -4 t cos (fa \/3 + 2 ). 

Thus, (7.410) and (7.411) represent possible vibrations of the 
particles. The most general vibration is given by adding these 
expressions, thus: 



(7.412) 



x = Ai cos (kt + 1) + Az cos (kt \/3 + 2 ), 

2/ = Ai COS (fa + i) ^2 COS (fa \/3 + 2). 



We know that this is the general solution, because it contains four 
constants of integration which may be chosen to satisfy initial 
conditions corresponding to given positions and velocities of 
the particles at t = 0. 
Let us suppose, for example, that when t = we have 



x = y = x = 0, 



v. 



This corresponds to the case where the motion is started by 
giving a blow to D. Putting t in (7.412) and the equations 
obtained by differentiating (7.412), we obtain 



(7.413) 



Ai COS 1 + Az COS 2 = 0, 
Ai COS 1 Az COS 2 = 0, 

sin 1 A* \/3 sin e 2 == 0, 



-Ai sin 1 + A 2 \/3 si 
which are four equations for Ai, A 2 , i, c 2 . 



v 
sm c 2 = T> 

A/ 

The solution is 



(7.414) l = 6 2 = fa Ai - - ^ ^2 
so that the motion of the particles is given by 



(7.415) 



r sin %2 -- ^= sin (Art \/3) h 
~ sin A; + ^ sin (A; \/3) J- 



This motion is complicated, but the simple harmonic motions 
of which it is composed are easy to describe. These simple 
harmonic vibrations are called normal modes of vibrations. They 



SBC. 7.4) MOTION OF A RIGID BODY AND OF A SYSTEM 211 

are executed by a vibrating system when the initial conditions 
are properly chosen. 

The vibration given in (7.415) is not a normal mode of vibra- 
tion, nor in general is that given by (7.412). But if the initial 
conditions are chosen so that A 2 = 0, we have the normal mode 
of vibration (7.410), whereas, if the initial conditions are chosen 
so that A i = 0, we have the normal mode of vibration (7.411). 
The periodic times and frequencies of normal modes are called 
normal periods and normal frequencies. In the above problem 
the normal periods are 

27T 27T 

fc' k V5 

When a system is executing a normal vibration, its configura- 
tions are usually simple to describe. Thus, in the mode (7.410) 
we have x = y, and in (7.411) we have x = y. Typical 
configurations for the normal modes are shown in Figs. 93a and 6. 




(6) 

Fio. 93. (a) Loaded string vibrating in first normal mode, 
vibrating in second normal mode. 



(6) Loaded string 



We have been discussing the vibrations of a particular system 
two particles on a taut string. All problems of vibration have 
certain features in common; and although we shall not attempt 
here to prove these facts, it will be useful to sum them up : 

(i) A vibration may be regarded as a superposition, or addi- 
tion, of simple harmonic vibrations. 

(ii) Each simple harmonic vibration is called a normal mode of 
vibration. It is possible to make a system vibrate in a normal 
mode by starting with suitable initial conditions. 

(iii) The periods and frequencies of the normal modes are 
called the normal periods and frequencies. 



212 PLANE MECHANICS [SEC. 7.4 



(iv) The number of normal modes is equal to the number of 
degrees of freedom of the system. 

(v) The normal periods are found by solving a determinantal 
equation, e.g., (7.406). 

Vibrations of a particle in a plane. 

As another example of a vibrating system with two degrees of 
freedom, let us consider a particle which moves in a plane in 
a field of force such that the potential energy per unit mass is 

(7.416) V = i(az 2 + 2hxy + by*), 

where a, hj b are constants. The force components per unit 
mass are then 

(7.417) X = -(ax + hy), Y = -(hx + by). 

These vanish at the origin, which is therefore a position of 
equilibrium. 

The equations of motion are 

(7.418) x = -ax - hy, y = -hx - by. 
Trying a solution 

(7.419) x = A cos (nt + e), y = B cos (nt + c), 
we see that (7.418) are satisfied provided A, B, n satisfy 

(7420) ( A(n* - a) - Bh = 0, 

^ } \ -Ah + B(n*-b) =0. 

Hence, n must satisfy the determinantal equation 

2 - a -h 

-h n 2 -b 

or 

(7.422) n 4 - n\a + 6) + ah - W = 0; 

the solutions are n\, HI, where 



(7.421) 



= 0, 



6) - V(a - 6) 2 + 4J. 

If one of these values should be negative, the corresponding n 
would be imaginary, and the solution (7.419) would contain 
hyperbolic instead of trigonometrical functions. This case will 



SEC, 7.4] MOTION OF A RIGID BODY AND OF A SYSTEM 213 

be discussed in Sec. 7.5; for the present, we assume that ni, n a 
are real. 

Then the normal periods are 2rr/ni, 2ir/n 2 , and the normal 
modes of vibration are 

vft * ct 

(7.424) x = Ai cos (nit + 1), y = -^r AI cos (nrf + i), 
and 

(7.425) x = A z cos (n z t + 2 ), y = ^-^ A 2 cos (n z t + e 2 ), 

where Ai, A 2, ei, c 2 are arbitrary constants. The general motion 
is found by adding the solutions (7.424) and (7.425), just as we 
added (7.410) and (7.411). 

The preceding discussion is actually more general than might 
appear. Let us again suppose that a particle moves in a plane 
under a conservative force system, with potential energy V per 
unit mass; but instead of assuming the simple expression (7.416) 
for y, we shall merely assume that it is a function which can be 
expanded in a Taylor series. 

Let Xo, 2/0 be a position of equilibrium. Since the components 
of force must vanish there, we have 






().-* 



the suffix zero indicating evaluation at x = XQ, y = y$. If we 
expand V in a Taylor series about x , 2/0, two terms in the expan- 
sion vanish on account of these equations, and so 



(7.427) F-F. + 

(y - ,). 



the terms not written being of a higher order of smallness if 
x x Q , y 2/0 are small. 

Now V is always undetermined to within an additive con- 
stant; there is therefore no loss of generality in putting V = 0. 
If we shift the origin to the position of equilibrium (z , ^o) and 
define a, A, 6 by 



214 PLANE MECHANICS [Sic. 7.5 



the principal part of 7 for small values of x and # is 
(7.429) V = i(az 2 + 2hxy + fey 2 ), 



which is formally the same as (7.416). The deductions based on 
(7.416) were exact; the same formal deductions hold approxi- 
mately for small vibrations about any position of equilibrium. 
The normal modes of vibration are given by (7.424) and (7.425), 
where ni, n 2 are given by (7.423), a, ft, 6 having the values (7.428). 

7.6. STABILITY OF EQUILIBRIUM 

A position of equilibrium for any system is said to be stable 
when an arbitrary small disturbance does not cause the system to 
depart far from the position of equilibrium. Otherwise, it is 
unstable. By "small disturbance" we mean that, at the initial 
instant, the particles of the system are displaced from their 
positions of equilibrium through small distances and their 
velocities are small. The system is stable if in the resulting 
motion the particles remain at small distances from their positions 
of equilibrium. 

Thus a compound pendulum hanging from its axis of support 
is in stable equilibrium. If it is balanced with its mass center 
above the axis of support, the equilibrium is unstable, because a 
small disturbance will cause the pendulum to move right away 
from the position of equilibrium. 

Condition of minimum potential energy. 

Let a system have a potential energy V. We know by the 
principle of virtual work (cf. Sec. 2.4) that, for a system in 
equilibrium, no work is done in a small displacement. Thus 
5V = for any small displacement from a position of equilibrium, 
and so V has a stationary value there. 

Stationary values are of various kinds; the question of stability 
turns on the character of the stationary value of 7. We make 
the following statement: //, in a position of equilibrium, the 
potential energy is a minimum, then the equilibrium is stable. 

To prove this, let us recall the principle of energy, 

(7.501) T + V jB, 



SBC. 7.5] MOTION OF A RIGID BODY AND OF A SYSTEM 215 

where E is a constant. Since potential energy is always undeter- 
mined to within an additive constant, there is no loss of generality 
in assuming V = at the position of equilibrium. Then, since 
V is a minimum there, we have V > for all positions near that 
of equilibrium. The constant E is found from the small initial 
disturbance. Let To, VQ be the initial kinetic and potential 
energies. Then E = To + F , which is small and positive. In 
the subsequent motion, 

(7.502) V = E - T < E, 

since T cannot be negative. Thus V always remains less than 
the small positive constant E, and so the equilibrium is stable, 
since to escape to a finite distance from the position of equilibrium 
the potential energy would have to become finite. 

As an illustration, consider a simple pendulum of mass m and 
length a. At an inclination to the downward vertical, the 
potential energy is 

V = mga(l cos 6), 

if we choose V = at the position in which the pendulum hangs 
vertically. Then V is a minimum for = 0. Suppose that the 
pendulum is disturbed to an angle and is given a kinetic energy 
To. In the subsequent motion, as in (7.502), 

mga(l cos 6) < To + mga(l cos ), 

where the right-hand side is small. Thus cos 6 must remain 
nearly equal to unity, or, in other words, must remain small. 

If, on the other hand, we consider that position of equilibrium 
in which the pendulum is balanced directly above the point of 
support (the string being replaced by a light rod) and measure 
from this position, we have 

V = mga(cu8 -1). 

Then V is a maximum for = 0. Our inequality (7.502) is still 
valid; it reads 

w0ra(cos 1) ^ To + mgra(cos 1), 

where 0o, TQ refer to the initial disturbance. But this inequality 
is not violated as increases from to ?r, and so it does not 



216 PLANE MECHANICS [SBC. 7.5 

restrict the motion. The position of equilibrium is actually 
unstable, but this inequality shows only that it may be so. 

The following statement is true for a system with any number 
of degrees of freedom, but we shall prove it here only for systems 

with one degree of freedom: // the 
potential energy at a position of 
equilibrium is not a minimum, then 
the equilibrium is unstable. * * 

Let x be the variable which fixes 
the position of the system. Con- 
sider the graph of the potential 




x energy V against x (Fig. 94). 

FIG. 94-Graph of potential B the pr i nc i p l e o f virtual WOrk, 

energy against position; stable ^ f r- j 

equilibrium at A and D, unstable WC have 8V = for an mfillltesi- 

equiiibrium at B and c. mal displacement dx f rom a position 

of equilibrium. In fact, at a position of equilibrium 



(7.503) - Q, 

so that the positions of equilibrium correspond to those points 
on the graph where the tangent is parallel to the z-axis, i.e., the 
points A y B, C, D. At A and D, V is a minimum, and hence we 
know that equilibrium at A or D is stable. 

Let us now consider the position corresponding to B. Suppose 
the system is displaced to a neighboring position B' and is then 
released from rest. Since the tangent at B' is not parallel to the 
#-axis, the system cannot remain in equilibrium at B'. It must 
start to move; and since its kinetic energy (being positive) must 
increase in comparison with its initial zero value, V must decrease, 
and so the system must move still farther away from B. It can 
come to rest only when V takes the same value as at B' . Thus it 
cannot stop moving until it has passed the position corresponding 
to A. Clearly, this is a case of instability. 

At C the potential energy has a stationary value, but it is 
neither a maximum nor a minimum. By considering an initial 
displacement in the direction of D, it is seen that the equilibrium 

* In the particular case where V is constant (as for a sphere resting on a 
horizontal table), the equilibrium is often called neutral. Actually, it is 
unstable in the sense of our definition. 



SEC. 7.5] MOTION OF A RIGID BODY AND OF A SYSTEM 217 

is unstable. This completes the proof of the italicized statement 
on the preceding page. 

To sum up : A position of equilibrium is stable if, and only if, the 
potential energy is a minimum. 

Hence, in the case of a system with one degree of freedom, a 
sufficient condition for stability is 



(7.504) > 0, 

at the position of equilibrium. This condition is also necessary, 
unless d*V/dx* = at the position of equilibrium; in that 
exceptional case, we have to examine the higher derivatives. 

It is clear from Fig. 94 that between any two positions of stable 
equilibrium there must be at least one position of unstable equi- 
librium. Points such as C, where a point of inflection on the 
graph coincides with a tangent parallel to the z-axis, are excep- 
tional. In general, positions of stability and instability alternate. 

Stability of equilibrium of a particle in a plane. 

It was shown in (7.429) that, near a position of equilibrium, the 
potential energy per unit mass of a particle in a plane may be 
written 

(7.505) V = $(ax* + 2hxy + by*), 

where a, h, b are constants. It is known, from the analytical 
geometry of conies, that we may choose new rectangular axes 
Ox'y' such that 

(7.506) ax* + 2hxy + by* - aV 2 + b'y'*, 

where a', b' are new constants. 
Let us recall how a', b f are found. For any constant value of X, 

(7.507) ax* + 2hxy + by* - \(x* + y*) 

= a'x'* + b'y'* - X(z' 2 + y'*). 

If X = a' or X = &', the right-hand side is a perfect square. For 
either of these values of X, the left-hand side must also be a 
perfect square. Thus, if X = a' or X = &', 

(7.508) (a - X)(6 - X) - h* = 0, 
or, in determinants! form, 

a X h 

h b - X 



(7.509) 



0. 



218 PLANE MECHANICS [Sue. 7.5 

In fact, a' and V are the roots of this quadratic equation. 

Suppose the transformation carried out, so that, near the 
position of equilibrium, 

(7.510) V - i(aV + 6V 2 ). 
The equations of motion are 

W 

x dx 7 * ~ aV ' 

(7.511) f 

'' 



or 

(7.512) x' + aV =0, #' + Vy' = 0. 

Various cases have now to be distinguished: 

(i) a! > 0, x' = A cos (Vo 7 + B sin (Vo 7 - 0, 

(ii) a' = 0, x' = A* + 5, 

(in) a' < 0, a;' = Ae^'^'-' + Ber'S^''*. 

These are the solutions of the first of (7.512), according to the 
sign of a! . The solutions of the second equation for y' may be 
similarly classified according to the sign of b'. 

The solution for case (i) indicates that x' remains permanently 
small, so that there is stability as far as x r is concerned. The 
solutions for cases (ii) and (Hi) indicate instability. Thus, there 
is stability if, and only if, both a' and b' are positive; since a', b 1 
are the roots of (7.509), we may state our result as follows: 

When the origin is a position of equilibrium, the potential energy 
for adjacent positions is given by (7.505). The equilibrium is 
stable if, and only if, the two roots of the determinantal equation 
(7.509) are positive. 

It is clear from (7.510) that V is a minimum at the origin if, and 
only if, the roots of (7.509) are positive. Hence, we have a direct 
proof in this case that minimum potential energy is the condition 
for stability, both necessary and sufficient. 

In the case of stability, oscillations along the axes of 
x' and y' are normal modes and the normal periods tiro 



where Xi, X 2 are the roots of (7.509). 



SBC, 7.5J MOTION OF A RIGID BODY AND OF A SYSTEM 219 



Problems of balancing. 

Theoretically it is possible to balance a needle on its point, but 
in practice it is extremely difficult to do so. There is a position 
of equilibrium with the needle vertical, but it is unstable. The 
instability is obvious in view of the general test given above, 
because the height of the center of gravity is decreased as the 
needle is displaced from the vertical position, and so the potential 
energy is a maximum for the vertical position. 






(a) 

FIG. 95. Cylindrical body rolling on a horizontal piano: (a) position of equi- 
librium, (b) displaced position. 

If, instead of a needle, we try to balance a body with a rounded 
base, it is not immediately evident whether the equilibrium in the 
position of balancing is stable or not. But the condition of 
minimum energy gives us an easy test. We shall confine our 
attention to cases where the possible motion of the body is 
two-dimensional. 

Figures 95a and b show end views of a cylindrical body in 
contact with a rough horizontal plane; the lower part of the 
section is a circular arc of radius a. In equilibrium (Fig. 95a) 
the center of gravity C must lie vertically above the point of 
contact, since the body is in equilibrium under two forces (the 
weight and the reaction) and their lines of action must coincide. 
Let h be the height of the center of gravity above the point of 
contact. 

Figure 956 shows a displaced position, in which the body has 
been turned through an angle 0. If W is the weight of the body, 
the potential energy is 

(7.513) V = W[a - (a - h) cos 0]. 

Hence, 



< 7 - 514 > 



w (0 ~ 



220 



PLANE MECHANICS 



[SEC. 7.5 
if the 



Thus the equilibrium is stable if, and only if, a > h, i.e. 
center of gravity lies below the center of the circle. 

Let us now consider a more general problem, which includes the 
preceding as a special case. Let A be a cylinder of any section, 
balanced on a fixed cylinder A', the contact being rough and the 
common tangent horizontal (Fig. 96a). C is the center of 
gravity of A. D, D' are the centers of curvature of the sections 
of the cylinders at the point of contact, and p, p' are the radii of 
curvature; the height of C above the point of contact is h. 





(a) (6) 

FIG. 96. One cylindct i oiling on another: (a) position of equilibrium, (b) dis- 
placed position. 

Figure 96& shows a displaced position, in which the point of 
contact has moved through a small angle 0' about D', and the 
line DC now makes a small angle 6 with DD'. Since we are con- 
cerned only with small values of and 0', it is legitimate to regard 
the sections in the neighborhood of the point of contact as circular 
arcs of radii p and p'. Then, by the condition of rolling, 

(7.515) p0 = p'0'. 

If W is the weight of A, the potential energy is 

(7.516) V = W[(p' + p) cos 0' - (p - h) cos (0 + 0')], 



SEC. 7.5] MOTION OF A RIGID BODY AND OF A SYSTEM 221 

or, since 0, 0' are small, 

(7.517) V - W[(p - h)(9 + O'Y - (p' + p)0' 2 ] + C, 

approximately, where C is a constant. By (7.515), this may be 
written 

(7.518) V = iW [( P - /i) (l + p ^) 2 - ( P ' + P)] + C. 
The condition that this shall be a minimum for 6' = is 

(7.519) (p - Ji) (l + ^y - (p' + p) > 0, 

or, equivalently, 

(7.520) h<- p ^~,- 

P T- P 

This is the condition of stability. On letting p' > , it reads 
/t < p, agreeing with the result established earlier. If on the 
other hand we let p <*> , we get h < p' as the condition for the 
stability of a body with a flat base balanced on a cylinder with 
radius of curvature p'. 

By means of the principle of energy, it is easy to find the period 
of small oscillations of a stable balanced system when disturbed. 
For example, the body shown in Fig. 95 has the potential energy 
given by (7.513). To convert to dynamical units, we put 
W mg, where m is the mass of the cylinder. Thus, when is 
small, the potential energy is approximately 

(7.521) V = mgh + $mg(a - 



Since, at any instant, the cylinder is turning about the line of 
contact, the velocity of the mass center is approximately hi; the 
angular velocity of the cylinder is 6. Hence (7.302) gives 

(7.522) \mWfr + i/0 2 + mgh + mg(a - h)0* = E, 

where 1 is the moment of inertia about a line through C parallel 
to the generators. If we write 

(7.523) F = h* + ^, 

and differentiate (7.522), we get 

(7.524) W + g(a - K)B = 0; 



222 PLANE MECHANICS [SEC. 7.6 

this gives a simple harmonic motion with period 

(7.525) r = 2irk/Vg(a - h). 

7.6. SUMMARY OF APPLICATIONS IN PLANE DYNAMICSMOTION 
OF A RIGID BODY AND OF A SYSTEM 

I. Moments of inertia. 

(a) Definitions: 



r? or / = / r 2 dm = J7J"p(z 2 + t/ 2 ) dxdydz] 



(7.601) / = 



(6) Devices for calculation: 

(i) Theorem of parallel axes (all moments of inertia follow 
immediately when those for axes through mass center 
are known). 

(ii) Theorem of perpendicular axes (for a plane distribution, 
the moments of inertia about axes perpendicular to the 
plane follow immediately when those for axes in the 
plane are known). 
(c) Standard results: 



Body 


Axis 


k* 


Hoop of radius a 


Perpendicular to plane at 


a* 


Circular disk of radius a 


center 
Perpendicular to plane at 


4a a 


Circular cylinder of radius a 


center 
Geometrical axis 


$a 2 


Rod of length 2a 


Perpendicular to rod at 


*a 2 


Rectangular plate of edges 2<z, 26 . . 
Sphere of radius o, ... 


center 
Through center parallel 
to edge 26 
Perpendicular to plate at 
center 
Diameter 


fa 2 
*(a 2 + & 2 ) 
8a 2 









II. Kinetic energy (T) and angular momentum (K). 

(a) Rigid body turning about fixed axis: 
(7.602) T = i/co 2 , h = /o>. 



Ex. VII] MOTION OF A RIGID BODY AND OF A SYSTEM 223 

(6) Rigid body in general plane motion: 

(7.603) T = |mg 2 + i/co 2 , h - / (h, I about mass center). 

III. Motion of a tigid body. 

(a) Rotation about fixed axis: 

(7.604) 7o> = N (angular momentum), 

(7.605) i/o> 2 + V = E (energy). 

(b) Compound pendulum: 

(i) Exact equation of motion: 

(7.606) P0 + ga sin = 0, (k relative to axis), 
(ii) Periodic time for small oscillations: 

2irk 

(7 ' 607) T = VTa 

(111) Equivalent simple pendulum: 

(7.608) I = ~ 

(c) General motion parallel to plane : 

(7.609) mx = X, my = 7, /w = N (momentum); 

(7.610) img 2 + i/w 2 +V = E (energy). 

IV. Normal modes of vibration. 

(a) A vibration is in general not periodic; it is composed of 
simple harmonic vibrations with different frequencies. These are 
the normal modes. A system vibrates in a normal mode if 
started under special initial conditions. 

(6) The normal frequencies are found by assuming a simple 
harmonic solution of the equations of motion and solving a 
determinantal equation obtained on this assumption. 

V. Stability of equilibrium. 

Necessary and sufficient condition for stability: the potential 
energy is a minimum. 

EXERCISES VH 

1. A uniform rod of length I and mass M is free to rotate in a vertical 
plane about an axis at a distance a from its center. If it is released from a 



224 PLANE MECHANICS [Ex. VII 

horizontal position, find its angular velocity when passing through the 
vertical position. 

2. A bucket of mass M is fastened to one end of a light rope; the rope 
is coiled round a windlass in the form of a circular cylinder (radius a) which 
is left free to rotate about its axis. Prove that the bucket descends with 
acceleration 

g 

1 + (//Jl/a 2 )' 

where / is the moment of inertia of the cylinder about its axis. 

3. Three uniform rodrf, each of mass w, form an equilateral triangle of 
side 2o. The triangle is suspended from one corner. Find the lengths 
of the equivalent simple pendulums for oscillations under gravity 

(i) when the triangle oscillates in its own plane; 

(ii) when the plane of oscillation is perpendicular to the plane of the 
triangle. 

4. A wheel consists of a thin rim of mass M and n spokes each of mass m, 
which may be considered as thin rods terminating at the center of the wheel. 
If the wheel is rolling with linear velocity v, express its kinetic energy in 
terms of M, m, n, v. 

With what acceleration will it roll down a rough inclined plane of inclina- 
tion a? 

5. A buoy is formed by joining the edge of a thin metal conical shell to 
the edge of a hemispherical shell of the same material and thickness. The 
radii of the hemisphere and of the mouth of the cone are each equal to 5 ft., 
and the slant height of the cone is 10 ft. The buoy is placed with the hemi- 
sphere in contact with the rough horizontal surface of a dock so that the 
axis is vertical. If slightly disturbed, determine whether or not it will 
return to the vertical position. 

6. One end of a heavy chain is attached to a drum and the chain is 
wrapped around the drum, making n complete turns, with a small piece of 
chain hanging free. The drum is mounted on a smooth horizontal axle, 
and the chain is allowed to unwrap itself. Apply the principle of energy to 
find the angular velocity of the drum at the instant when the chain is com- 
pletely unwrapped, in terms of the mass of the chain (m}, the radius of the 
drum (r), and the moment of inertia of the drum (7). 

7. A rectangular plate swings in a vertical plane about one of its corners. 
If its period is 1 sec., find the length of the diagonal. 

8. A particle of mass m moves in a plane under the action of a force 
with components 

X - -k*(2x + y), Y - -*(x + 2y), 

where k is a constant. What is the potential energy? Find the normal 
periods of oscillation about the position of equilibrium. 

9. A uniform circular plate of radius a and mass M is dragged along a 
smooth sheet of ice by means of a long string attached to a point A on the 



Ex. VII] MOTION OF A RIGID BODY AND OF A SYSTEM 225 

rim of the plate. The tension T in the string is kept constant throughout. 
If initially the plate is at rest and the diameter through A makes a small 
angle with the string, show that this diameter oscillates about the direction 
of the string with a period equal to 



27'" 

10. A particle A hangs from a fixed point by a light string, and another 
particle H of the same mass hangs from A by a second light string of the 
same length. Find the normal periods of oscillation, and sketch the normal 
modes 

11. A homogeneous solid cylinder, whose section is a semicircle of radius a, 
rests with its flat face horizontal and in contact with a fixed rough circular 
cylinder of radius 6, the generators of the two cylinders being parallel. 
Find the greatest value of a/6 for which there is stability. 

12. Find the radius of gyration of a uniform semicircular plate about a 
line through the mass conter perpendicular to the plate. 

13. A ladder (length 2a) rests against a smooth vertical wall and a smooth 
horizontal floor, the inclination to tho floor boing initially . Find the 
inclination of the ladder to the floor at the instant when the upper end 
leaves the wall as it slides down under the action of gravity. 

14. A pendulum consists of a bob of mass m at the end of a light rod of 
length 3a. It is suspended from the point of the rod distant 2a from the 
bob. A horizontal force mb cos nt (where b and n are constants and 6 is 
small) is applied to the rod at its upper end. Find the angular amplitude 
of the forced oscillations of period 2ir/n. 

15. A uniform solid ellipsoid of revolution of semiaxes a, b (the axis of 
revolution boing 2a) is cut in two by a plane through the conter perpendicular 
to the axis of revolution. If either half will balance in stable equilibrium 
with its vertex on a horizontal piano, prove that 



16. When a ship rolls through a small angle, the upward thrust of the 
water intersects the central plane of the ship at a point called the metacenter. 
Find a formula for the periodic time of rolling of a ship in terms of /t, the 
height of the metacenter above the mass center of the ship, and /r, the radius 
of gyration of the ship about a fore-and-aft axis through the mass con tor. 

Is the periodic time increased or decreased by shifting cargo horizontally 
from the center of the ship to the sides, this shift being done symmetrically 
with respect to the central plane of the ship? 

17. A square frame, consisting of four equal uniform rods of length 2a 
rigidly joined together, hangs at rest in a vertical plane on two smooth 
pegs Pj Q at tho same lovel. If PQ c and the pegs are not both in contact 
with the same rod, show that there arc three positions of equilibrium, pro- 
vided a < c \/2. 



226 PLANE MECHANICS [Ex. VII 

Of these positions, show that the only unstable one is the symmetrical 
position. If, however, a > c -\/2, show that the only possible position of 
equilibrium is stable. 

18. A particle is suspended by a light string of length a from the lower 
end of a rod of the same mass and length 2a, which is free to turn about its 
upper end. For vibrations about equilibrium in a vertical plane, show that 
the two normal frequencies arc given by 



where p satisfies 

4p 2 - 25p + 9 = 0. 

19. A rod of length 2a hangs from a support which is given a small hori- 
zontal displacement varying with time according to the equation 
= 6 sin pt, where 6 and p are constants. Find the equation of motion 
for small oscillations. Integrate the equation, obtaining a result with two 
arbitrary constants. Find these constants on the assumption that when 
t * the rod is hanging vertically and has no angular velocity; hence 
show that the inclination of the rod to the vertical is given by 



(n sn pt - p sn 



20. Two simple pendulums, each of mass m and length a, hang from a 
trolley of mass ^l/ which can run without friction along horizontal rails. 
A small impulse, parallel to the rails, is applied to one of the pendulums and 
imparts to it an angular velocity o>, the other pendulum and the trolley 
having no velocity at that instant. Investigate the resulting motion, and 
express the displacement of the trolley and the inclinations of the pendulums 
to the vertical as functions of the time. 

Show that, if the ratio m/M is small, the motions of the pendulums 
relative to the trolley may be regarded as simple harmonic motions with 
slowly varying amplitudes, the amplitudes being given by the absolute 
values of 



where 



,... 



CHAPTER VIII 
PLANE IMPULSIVE MOTION 

8.1. GENERAL THEORY OF PLANE IMPULSIVE MOTION 
The concept of an impulsive force. 

For a particle moving in a plane under the action of a force 
with components X, Y, the equations of motion are 

(8.101) mx = X, my = Y. 
Multiplying by dt and integrating from t to t\, we obtain 

(8.102) AM) - Fxdl, A(my) = f" Y dt, 

/fo Jt9 

where A denotes an increment during the time interval (fo, ti). 
The vector with components 

(8.103) rXdt, C tl Ydt 

JlQ /'o 



is called the impulse on the particle during the time interval 
(to, ti). We may state (8.102) in words as follows: The increment 
in momentum is equal to the impulse. 

Let us now suppose that a particle of mass m can move along 
the x-axis. At time t = 0, it is at rest at x 0. At this 
instant a force 

(8.104) X = A sin ^ 

commences to act and acts until t = T. (A and r are constants.) 
During this time the equation of motion of the particle is 

(8.105) mx A sin > 
and so 



Ar A ;rA 

U = X = ( 1 COS I 

irm\ r) 



Art AT* . irt 

_ sm _, 

irm ?T 2 m r 

227 



(8.107) X = 



228 PLANE MECHANICS [SBC. 8.1 

the constants of integration having been chosen to fit the initial 
conditions. Thus in the time interval (0, T) the particle receives 
increments in velocity and position given by 

/O m*N A 2 ^ T A AT * 

(8.106) Aw = > Az = 

v ' TTW TTW 

The mean value of the force is 

f r A ' fit 

T Jo r TT 

and so (8.106) may be written 

V" V^ 2 

(8.108) AM = , Az = 
v m m 

The experiment may be repeated using different values of 
A and T. We note that as long as the product XT remains 
unchanged the value of Aw remains the same. If we let A (and 
therefore X) tend to infinity and let T tend to zero in such a way 
that XT maintains a fixed value (7, we have 

(8. 109) AM -> , Ax - 0. 

We note that, in this limiting case of "an infinite force acting for 
an infinitesimal time," there is an instantaneous change in 
velocity but no change in position. 

Returning to the general equations (8.102), we may let the 
force components (X, Y) tend to infinity and the time interval 
ti fo to zero in such a way that the integrals remain constant or 
approach finite limits. Under these circumstances a particle 
moving in a plane experiences (in the limit) an instantaneous 
change of velocity. Since the velocity remains finite during this 
change, the displacement is zero in the limit. The instanta- 
neous change in momentum is given by 

(8.110) A(mz) = lim P 1 X dt, A(m#) = lim P 1 Y dt. 



Have we here introduced a new ingredient or concept into 
mechanics? It must be admitted that we have, because no 
force, however large, can produce an instantaneous change in 
momentum. To place our new ideas on a secure foundation, we 
admit the concept of an impulsive force, with components denoted 



SEC. 8.1] IMPULSIVE MOTION 229 

by X, P: it is such that, when applied to a particle, the impulsive 
force causes an instantaneous change in momentum given by 

(8.111) A(mx) = X, A(my) = f. 

We must, however, regard the impulsive force, not as something 
absolutely new, but as connected with the ordinary force (X, Y) 
by the relations 

(8.112) X = lim r X dt, Y = lim P Y dt, 

ti-*to J to ti-+to Jh 



obtained by comparison of (8.110) and (8.111). 

On account of this connection, it is unnecessary to repeat for 
impulsive forces results already obtained for ordinary forces. 
We draw attention to the fact that impulsive reactions between 
the particles in a rigid body obey the law of action and reaction. 
The theory of moments applies to impulsive forces, and we may 
speak of an impulsive couple. The idea of equipollence may also 
be used. 

It is clear from (8.112) that forces which remain finite as 
ti to (e.g., gravity) contribute nothing to the impulsive force. 

An impulsive force is the product of an ordinary force and a time and is 
equal to a change in momentum. Hence, impulsive force has the dimensions 
[MLT* 1 ]; its magnitude is expressed in dyne sec. or gm. cm. sec." 1 in the 
c.g.s. system and in poundal sec. or Ib. ft. sec." 1 in the f.p.s. system. 

Principles of linear and angular momentum. 

We expressed in (5.206) the law that, for any system, the rate 
of change of linear momentum is equal to the sum of the external 
forces. If M x , My are the components of linear momentum in 
the directions of axes Ox, Oy, and X, Y the total components of 
external force in those directions, then 

(8.113) A, = X, M v = Y. 

Let us multiply by dt, integrate from t = fc to t = ti, and then 
proceed to the limit t\ > fo, supposing the forces to tend to 
infinity. Then 

(8.114) AM* = X , AM tf = Y, 

where X, Y are the sums of the components of the external 
impulsive forces. In words, the sudden change in the linear 
momentum of a system is equal to the total external impulsive force. 



230 PLANE MECHANICS [Sue. 8.1 

These equations may also be written in vector form: 

(8.115) AM = F, 

where F is the vector sum of the external impulsive forces. 
Similarly, we obtain from (5.209) the vector equation 

(8.116) mAq = F, 

where m is the mass of the system and Aq the sudden change in 
the velocity of the mass center. 

If there are no external impulsive forces, we have F = 0, and 
hence Aq = 0. Thus, when the impulsive forces are purely 
internal, there is no sudden change in the velocity of the mass 
center. 

This result is of interest in connection with collisions and 
explosions. Here, in physical reality, we find large forces acting 
for short intervals of time, and we may treat the phenomena 
mathematically by means of impulsive forces. Thus, if a 
shunting locomotive strikes a car, the mass center of the system 
"locomotive + car" has the same velocity just before and just 
after the collision. The bursting of a shell in the air produces a 
set of fragments, the mass center of which has the same velocity 
as the mass center of the shell before bursting. 

Consider now the change in angular momentum about a fixed 
line due to the action of impulsive forces. Equation (5.214) 
applies. Multiplying by dt, integrating over the range (to, ti) 
and proceeding to the limit as usual, we find 

(8.117) M = lim f tl Ndt = ft. 

i-*fo Jt* 

In words, the sudden change in angular momentum about the fixed 
axis is equal to the moment of the external impulsive forces about 
the axis.* 

We may treat similarly the equation (5.219) which concerns 
motion relative to the mass center. We find that the sudden 
change in angular momentum relative to the mass center is equal 
to the moment of the external impulsive forces about the mass center. 

If the system is a rigid body with a fixed axis, (8.117) may be 
written 

(8.118) 7A-#, 

* It is easily proved that, $, defined as the limit of the time integral of the 
moment, is equal to the moment of the impulsive forces. 



SBC. 8.2] IMPULSIVE MOTION 231 

where 7 is the moment of inertia about the fixed axis, Aw the 
change in angular velocity, and N the moment of the impulsive 
forces about the axis. 

For a rigid body which can move parallel to a plane, (8.118) 
holds, provided we understand 7 to be the moment of inertia 
about the mass center and N the moment of impulsive forces 
about the mass center. 

We have now converted the principles of linear and angular 
momentum into forms valid in the case where impulsive forces 
act. Since the application of impulsive forces leads to sudden 
changes in velocity, we may call the theory of impulsive motion 
a discontinuous theory, reserving the word continuous for those 
cases in which no sudden changes in velocity occur. 
1 In the continuous theory the principle of energy is useful for 
determining motions, either completely or in part. But it must 
be used with great caution in the discontinuous theory, because 
we find in general that when impulsive forces act the law of 
conservation of mechanical energy does not hold. Actually, 
the energy is not lost; it is converted into heat or employed to 
deform the bodies on which the impulses act. But mechanical 
energy disappears, and it is with mechanical energy alone that 
we are concerned in this book. 

Exercise. An impulsive force P is applied at one end of a bar of mass m 
and length 2o, in a direction perpendicular to the bar. Find the velocity 
imparted to the other end of the bar, assuming (i) that the center of the 
bar is fixed, (ii) that the bar is free. 

'8.2. COLLISIONS 

As remarked above, a collision between two bodies gives rise 
(in physical reality) to large reactions acting for a short time, 
and so we treat the problem of collision mathematically by means 
of impulsive forces. 

The collision of spheres and the coefficient of restitution. 

As an illustrative example, we shall discuss the problem of 
the collision of two spheres which are moving along the line 
joining their centers (Fig. 97). 

Taking the axis Ox along the line of centers, let us use the 
following notation: 



232 PLANE MECHANICS [Sue. 8.2 

mi, m z = masses of the spheres, 
Ul) uz velocities of centers before collision, 
u i> u * ^ velocities of centers after collision, 

P = magnitude of impulsive reaction. 
We have then 

(8.201) mi(u{ - ui) = -P, m 2 (u' 2 - u 2 ) = P, 
and so 

(8.202) miu{ + m 2 u f z = miUi + m 2 u 2 , 

as indeed we might have deduced directly from the fact that 
there is no external impulsive force. 

Our problem is to find the result of the collision, i.e., to find 
u{, u't when ui, u 2 are given. But for this we have only one 
equation (8.202), and that is not enough to give two unknowns. 
We can proceed no further without an additional hypothesis, 
and here we introduce the idea of the coefficient of restitution. 




FIG. 97. Collision of two spheres. 

Consider the problem of collision as it might occur in reality, 
say between two tennis balls. Actually the balls would become 
distorted during the collision and then would bound away from 
one another, regaining their spherical shapes. This is a com- 
plicated process which we cannot follow through mathematically, 
and we are obliged to substitute some simple hypothesis based 
on experimental results. 

We introduce the expressions speed of approach q a and speed of 
separation q 8 . For a general collision, these speeds are calculated 
for the particles of the two bodies at the point of contact, com- 
ponents of velocity along the common normal at that point being 
used. In our problem of colliding spheres, we have 

(8.203) q a = HI u 2 , q 8 = 14 u{. 



SEC. 8.2] IMPULSIVE MOTION 233 

The following general hypothesis is adopted: The speeds of 
separation and approach are connected by the relation 

(8.204) q 8 = eq a , 

where e is a positive number, called the coefficient of restitution. 
The value of e depends on the materials of which the bodies are 
composed and also on their shapes and sizes; it never exceeds 
unity in value. When e = 1 the bodies are said to be perfectly 
elastic, and when e = they are said to be perfectly inelastic. 
In the problem of the spheres, we now have 



(8 205) 

2 = i 2 . 

Hence, 

, _ mi -em* + (l+e} _J2_ 

} 



(8.206) 



t /i i \ wii . mz em\ 

u' 2 = (1 + e) - : - ui H -- - : - u z , 
mi + m 2 mi + m z ' 



and so the problem is solved. 

In the case of perfectly inelastic spheres (e = 0), we have 
HI = u' z ; there is no rebound. 

If the spheres are of the same mass (mi = m 2 ) and there is 
perfect elasticity (e = 1), we have 

(8.207) u{ = u Zf u' 2 = m; 

this means that the spheres exchange velocities. This case is 
particularly interesting because, in the kinetic theory of gases, 
the mathematical model represents the molecules by perfectly 
elastic spheres. 

Compression and restitution. 

The hypothesis (8.204) appears artificial; we generally prefer 
to adopt hypotheses which have some plausibility. The hypoth- 
esis may however be put in another form, which suggests 
rather better its connection with physical reality. To do this, 
we return to the physical picture of the collision of two tennis 
balls. At first the centers of the balls are approaching one 
another, and the balls are being distorted. Then they start to 
regain their spherical shapes, pressing against one another until 
they separate. Thus the whole period of collision is divided 



234 PLANE MECHANICS [Sac. 8.2 

into a period of compression and a period of restitution. We 
may adopt, instead of (8.204), the following hypothesis: The 
impulse during restitution bears to the impulse during com- 
pression a definite ratio e. Or, passing to the limit of infinite 
forces and vanishing time, we may make the following formal 
statement of our hypothesis: If Pi is the magnitude of the impul- 
sive reaction of compression required to reduce the speed of approach 
to zero, then the magnitude P 2 of the impulsive reaction of restitution 
is 

(8.208) P 2 = cPi 

where e is the coefficient of restitution. These two impulsive reactions 
act in the same direction. 

It is by no means obvious that (8.208) is equivalent to (8.204), 
but it can be proved without much difficulty. We shall here 
merely establish the equivalence for the problem of the spheres. 
We have, for compression, 

(8.209) m\u m\u\ = Pi, w 2 n w 2 w 2 = Pi, 



where u is the common velocity when the speed of approach is 
zero. For restitution, we have by (8.208) 



(8.210) miu'i m\u = ePi, 

We have here four equations, which can be solved for u(, u' 2 , u, PI. 
To prove that we get the same result as that given by (8.204), 
we eliminate u from (8.209) and also from (8.210) ; we obtain 

(8.211) / mim 2 (ui - u) = (mi + m 2 )Pi, 
\ WiW 2 (i4 ~ u() = e(mi + w 2 )Pi, 



from which (8.204) follows at once. 

Motion relative to the mass center. 

The mathematics of discontinuous motions is much simpler 
than that of continuous motions, because the equations to be 
solved are algebraic, not differential. But the algebra may 
become complicated, and it is sometimes advisable to use a 
special Newtonian frame of reference. Thus, in the case of the 
two spheres considered above, we may use a frame of reference 
in which the mass center is at rest before collision. It is, of 
course, at rest in this frame after collision also. We have then 



SBC. 8.3] IMPULSIVE MOTION 235 

(o.ZiL^) ' 



hence, 

(8.213) u{ = -etii, t4 = 

As a result of the collision, the components of velocity are 
reversed in sign and multiplied by the coefficient of restitution. 

If the kinetic energy is T before collision and T r after collision, 
we have 



O 



and so the loss of kinetic energy is 
(8.214) T - T' = (1 - e 2 )T. 

Since e ^ 1, kinetic energy is lost in every case except that of 
perfect elasticity (e = 1). 

We have discussed the collision of spheres in the case where 
their centers move along the line joining the centers. Provided 
the spheres are smooth, the extension to the case where the 
spheres have general motions is immediate; the 
components of momentum (and hence velocity) 
in directions parallel to the common tangent 
plane of the spheres undergo no changes, and 
the components of velocity along the common 
normal change as described above. 

8.3. APPLICATIONS 

We shall now illustrate the application of the 
principles of linear and angular momentum by 
two examples. 

The ballistic pendulum. 

Consider a rigid body, hanging in equilibrium 



N 
from a horizontal axis (Fig. 98). A bullet, 

traveling horizontally, strikes the body at A and FlQp 93. Baiiis- 
becomes embedded in it. As a result of the tic pendulum, 
impact, the body swings as a compound pendulum, rising through 
an angular displacement a before coming to rest. On account of 
its importance in ballistics, the apparatus is called a ballistic 
pendulum. From the angle a and the constants of the system, 
we can compute the velocity of the bullet, as we shall now show. 



236 PLANE MECHANICS [SEC. 8.3 

Let us take as dynamical system the body and the bullet. 
Then the forces between the body and the bullet are internal. 
During the brief interval of impact the only external forces 
acting are (i) gravity and (ii) the reaction at O. The force of 
gravity is a finite force and so contributes no impulsive force. 
Since the reaction at has no moment about 0, it is evident that 
the principle of angular momentum enables us to state that 

angular momentum of system about before impact 

= angular momentum of system about after impact. 

Let ON be the vertical through 0, AN being horizontal. 
Let m be the mass of the bullet, q its speed, / the moment of 
inertia of the body about 0, and w the angular velocity immedi- 
ately after impact. Then the angular momentum of the system 
about is 

before impact: mq ON, 

after impact: (m - AO 2 + ^)w- 

If the mass of the bullet is very small in comparison with that of 
the body, we may neglect m AO 2 in comparison with /; thus we 
have 

(8.301) mql = Io>, 

where I = ON. 

Although we cannot apply the principle of energy during 
impact, we can apply it in the subsequent motion. Thus, 
again neglecting the mass of the bullet in comparison with that 
of the body, we have 

(8.302) i/o> 2 = Mgh(l - cos a), 

where M is the mass of the body and h the distance of its mass 
center from 0. Hence, from (8.301), 



(8.303) 9 = -, 

which gives the speed of the bullet in terms of a and constants 
of the system. 

Linked rods. 

Two uniform rods AB, BC, each of mass m and length 2o, 
are connected by a smooth joint at B and lie in one straight line 



SEC. 8.3] IMPULSIVE MOTION 237 

on a smooth horizontal table (Fig. 99a). A horizontal blow P 
is struck at C, in a direction perpendicular to BC. We wish to 
find the motion generated. 

Let us draw a schematic diagram (Fig. 996), separating the 
rods in order to represent the reactions without confusion. 

ABC 



r 

FIG. 99o. A pair of rods, linked at B, receive a blow at f. 



y 








XN. 




A 


. _ f Y . . 
v,t^\ M >$ 1- v 2 f 
> c B > 


f c 




HI ^ r i 


- t 



o 

FIG. 99&. Diagram of velocities and impulsive forces. 

Taking rectangular axes Oxy, with Ox parallel to ABC and Oy 
in the sense of P, we shall use the following notation : 
u\, Vi = components of velocity of center of AB, 
u 2 , v z = components of velocity of center of BC, 
a)i angular velocity of AB, 
co2 = angular velocity of BC, 
X, F = components of reaction on BC at B, 
X, Y components of reaction on AB at B. 
Since the rods are joined at B, this point must have the same 
velocity whether considered as a point of AB or of BC. Thus, 

(8.304) u\ = u 2 , Vi + ctcoi = Vz ttoj2. 

The principle of linear momentum applied to each rod gives 

/ _ v v 

/r> O/\K\ J WM/i -~".A. frlUs .A. 

(8.305) < = _y = y 4- P- 

the principle of angular momentum gives 

(8.306) mk 2 o>i = -aY, mWu* -aF + aP, 



where k is the radius of gyration of each rod about its center, 
so that k* = a 2 . 



238 PLANE MECHANICS [Sac. 8.4 

From the first equation of (8.304) and the first two of (8.305), 
we see that 

(8.307) ui = u 2 = 0, 5 = 0. 

There now remain in (8.304), (8.305), and (8.306) five equations 
for the following five unknowns: 

01> 02, Wl, &>2j Y . 

It is most symmetrical to find f first by substitution in (8.304) 
from the other equations; we find 

(8.308) Y - iP, 
and hence 



(8.309) 



iP p 

0i=-i-, *-* S ^ 

PP 
___ o _f . 

^ ma T ma 



The velocity of is P/w, downward in the diagram. 

8.4. SUMMARY OF PLANE IMPULSIVE MOTION 

I. Components of impulsive force. 

(8.401) X = lim f tl Xdt, Y = lim f l 7 <B. 

<i-*fo^ a tv-*kJk 

II. Instantaneous change in motion, 
(a) Particle: 

(8.402) A(mx) = X, A(my) = f. 
(6) Any system: 

(8.403) AM* = X, &M V = K, A^ = N. 

(X, Y,N = total components and moment of external impulsive 

forces.) 

(c) Rigid body with fixed axis: 

(8.404) 7 Ao> = N. 

(d) Rigid body moving parallel to a fixed plane: 

(8.405) m Aw = , m A0 f , / Ao> & 

(w, v = components of velocity of mass center; N = impulsive 

moment about mass center.) 



Ex. VIII) IMPULSIVE MOTION 239 

III. Collisions. 

Either 

(8.406) q. eq a 
or 

(8.407) P 2 = ePi. 

(e = coefficient of restitution; e ^ 1.) 

EXERCISES Vin 

1. A bar 2 ft. long, of mass 10 lb., lies on a smooth horizontal table. It 
is struck horizontally at a distance of 6 in. from one end, the blow being 
perpendicular to the bar; the magnitude of the blow is such that it would 
impart a velocity of 3 ft. per sec. to a mass of 2 lb. Find the velocities of 
the ends of the bar just after it is struck. 

2. A uniform rod of mass m and length 3a hangs from a pin passing 
through it at a distance a from the upper end. Find in terms of m, a, g the 
magnitude of the smallest blow, struck at the lower end of the rod, which 
will make the rod describe a complete revolution. 

3. A ball is dropped on the floor from a height h. If the coefficient of 
restitution is e, find the height of the ball at the top of the nth rebound. 

4. A bar, 6 ft. long, is swinging about a horizontal axle passing through 
it at a distance of 1 foot from one end. At what point must a blow be 
struck to bring it to rest without causing any impulsive reaction on the 
axle? (This point is called the center of percussion.) 

5. A particle moving with a speed of 30 feet per second in a direction 
making an angle of 60 with the horizontal strikes a smooth horizontal 
plane and rebounds, the coefficient of restitution being $. Find the speed 
and the direction of motion of the particle immediately after impact. 

6. A uniform square plate of mass M and side 2a rests on a smooth 
horizontal table. A horizontal impulsive force of magnitude P is applied 
at a corner in a direction perpendicular to the diagonal at that corner. 
Show that the angular velocity generated by this impulsive force is 

3 



2Ma 

7. A tug of mass m tons is attached to a barge of mass M tons by a cable 
the mass of which may be neglected. The cable is slack. The tug moves 
and has acquired a speed of v ft. per sec. when the cable becomes taut and 
the barge is jerked into motion. Assuming that the cable has a coefficient 
of restitution and neglecting the impulsive resistance of the water, find 

(i) the speed imparted to the barge; 

(ii) the mean tension (in tons wt.) in the cable during the jerk, supposing 
this to take t sec. 

8. A billiard ball of radius a and mass M rests on a horizontal table. 
In a vertical plane through the center of the ball there is applied a horizontal 



24Q PLANE MECHANICS [Ex. VIII 

impulsive force of magnitude P. If the line of action of the impulse is at a 
height h above the table, find the initial velocity of that point of the ball 
which is in contact with the table. 

9. Two gear wheels of radii ai, a 2 and axial moments of inertia Ji, /2, 
respectively, can rotate freely about fixed parallel axles. Initially the 
wheel of radius ai is rotating with angular velocity w, while the other wheel 
is at rest. If the gear wheels are suddenly engaged, find the angular velocity 
of each wheel afterward. 

10. A beam of mass 100 Ib. and length 6 ft. hangs from an a*xle passing 
through it at a distance of 1 ft. from one end. It is drawn aside through 
an angle of 30 and then released. It is stopped dead at the lowest point 
of its swing by a horizontal blow which strikes it at a height of 2 ft. above 
its lower end, What is the impulsive reaction on the axle, expressed in Ib. 
ft. sec.- 1 ? 

11. Two uniform rods AB, BC, each of mass M and length 2a, are smoothly 
jointed together and rest on a smooth horizontal plane, the angle between 
the rods being 45. A horizontal impulsive force P is applied at A in a 
direction at right angles to AB and away from the rod BC. Find the initial 
angular velocity of BC. 

12. A smooth rod of length 2a and mass M rests on a horizontal plane. 
A small body of mass m moves in this piano with velocity v in a direction 
inclined to the rod at an angle of 45; it strikes the rod at a point distant c 
from the center. If the coefficient of restitution between the rod and the 
body is e, find the angular velocity of the rod and the velocity of the body 
after collision. 

13. A flywheel whose axial moment of inertia is 200 Ib. ft. 2 rotates with 
an angular velocity of 300 revolutions por minute. Find in ft. Ib. wt. sec. 
the angular impulse which would be required to bring the flywheel to rest. 
Hence find the frictional torque at the bearings if the flywheel romes to 
rest in 10 minutes under friction alone. 

14. One end of each of four equal uniform rods is smoothly jointed to the 
circumference of a uniform disk of radius a and mass M. The length of a 
rod is 2a and its mass is m. The points of attachment are at equal angular 
intervals. Initially the system is at rest on a smooth horizontal plane with 
each rod lying along a radius of the disk produced. A horizontal impulsive 
force of magnitude P is applied to the outer end of one rod in a direction 
perpendicular to it. Show that the initial angular velocity of the disk is 



a(M + 2m)' 

in a sense opposed to the direction of the impulse. 

15. A uniform rod of mass m and length 2a is moving on a smooth hori- 
zontal plane. At a certain instant, its center has velocity components u 
along the rod and v perpendicular to it, and the rod has an angular velocity o>. 
What impulsive force must be applied to a point of the rod at a distance b 
from the center in order to bring that point to rest instantaneously? 

16. Two uniform circular plates A and B t each of radius a and maso m, 
are connected by a rod of length 2a and mass m, each end of which is linked 



Ex. VIII] IMPULSIVE MOTION 241 

smoothly to a point on the circumference of one of the plates. The system 
is at rest on a smooth horizontal plane with the centers of the plates in the 
line of the rod produced. An impulsive couple $ acts on the plate A. 
Determine the initial motion of the plate B. 

17. AB t EC, CD are three equal rods, smoothly hinged to one another 
at B and C. They lie on a smooth horizontal plane, forming three sides of 
a square. AB can turn freely about A, which is fixed. An impulsive force 
applied to D sets D in motion with a velocity v directed away from A. 
Prove that the initial velocity of B is opposite in direction to that of D and 
equal in magnitude to -f^v. 

18. For the collision of two smooth latninas moving in a plane, prove that 
the assumption that the impulsive reaction of restitution is equal to e times 
the impulsive reaction of compression loads to the result that the ratio of 
the speeds of separation and approach is e. 

19. On a straight line /, there are situated n particles all of the same 
mass. Initially the particles are at the points A\, A 2t - - A n where 
OAi < OAz < - < OA n , O being a fixed point of L, and the velocity of 
the rth particle is in the direction OA r and of magnitude, u r . 

If ui > HZ > - > u n and the particles arc all perfectly elastic, find the 
final velocity of each particle. What would be the result if all the particles 
were perfectly inelastic,? 

20. A number of equal uniform rods are smoothly jointed together to 
form a chain which hangs at rest under gravity. The upper end A of the 
chain is free to slide on a smooth horizontal axis. If aii impulsive force is 
applied to A along the axis, show that the initial angular velocities of the 
last three rods are in the ratios 11 : 3: 1. 



PART II 
MECHANICS IN SPACE 



CHAPTER IX 
PRODUCTS OF VECTORS 

Up to the present, our development of mechanics has been 
restricted, for the most part, to two dimensions. We now come 
to the systematic treatment of mechanics in space. Here we 
must make a decision as to notation. On the one hand, we have 
the ordinary notation of coordinates; on the other hand, the 
vector symbolism. Each has its advantages, but on the whole 
the vector notation has proved more useful on account of its 
compactness. We shall therefore use it extensively (but not 
exclusively) throughout the rest of the book. The present 
chapter, together with Sec. 1.3, explains the mathematical 
language to be employed later. 

9.1. THE SCALAR AND VECTOR PRODUCTS 

In developing the theory of vectors, we try to extend to 
vectors the operations of ordinary (scalar) algebra, as far as 
possible. In Sec. 1.3, this was done successfully for the addition 
and subtraction of vectors and for the multiplication of a vector 
by a scalar. We now consider the multiplication of vectors by 
one another, and here the methods of ordinary algebra are not 
so easy to generalize. Actually, we define two types of product 
the scalar product and the vector product. 

As in Sec. 1.3, we use Pi, P 2 , Pa to denote the components of 
a vector P on rectangular axes Ox, Oy, Oz, and P to denote its 
magnitude. 

Scalar product. 

The scalar product of two vectors P and Q, written P Q, 
is defined by 
(9.101) P-Q = PQcosfl, 

where is the angle between P and Q. Since Q cos 6 is the 
component of Q in the direction of P (cf. Sec. 1.3), it is clear 
that P Q is equal to the magnitude of P multiplied by the 
component of Q in the direction of P. 

245 



246 MECHANICS IN SPACE [Sue. 9.1 

In particular, 3t P is the component of a vector P in the 
direction of a unit vector ^. Thus the work done by a force P 
in an infinitesimal displacement 3. 5s is, by (2.401), 

6W = P & 8s. 

Since the direction cosines of P are Pi/P, P*/P, P*/P and 
those of Q are Qi/Q, Q 2 /Q, Qs/Q, we have 



Hence, using (9.101), we have the following expression for the 
scalar product of two vectors in terms of their components: 

(9.102) P Q = PiQi + P.Q Z + P 3 Q S . 

From the definition, it is clear that the scalar product of two 
perpendicular vectors vanishes. 

Either from the definition (9.101) or from (9.102), it follows 
that the order of the factors in a scalar product is immaterial. 
Thus, 



in fact, scalar multiplication is commutative. It is also dis- 
tributive; that is, 

P.(Q + R) = P.Q + P-R. 

To show this, we recall that the components of Q + R are 
Qi + Ri, Q 2 + #2, Qs + Rs', and therefore, by (9.102), 

P (Q + R) = /MQx + R,) + P 2 (Q 2 + ,) + P 3 (Q 3 + fi.) 

= (PxQi + P 2 Q 2 + P 3 Q 3 ) + (Pi#i + P 2 # 2 + P 8 # 3 ) 
= P Q + P R. 

A third law governing the operation of multiplication in 
ordinary algebra, namely, the associative law, does not concern 
us here since we attach no meaning to P Q R. However, we 
have defined such quantities as (P Q)R and P(Q R), each 
being the product of a vector by a scalar. These quantities 
are, of course, quite different, one being a vector with the direc- 
tion of R and the other a vector with the direction of P. 

Exercise. A vector has components (1, 3, 2) in the directions of rec- 
tangular axes Oxyz. What is its component along the line x y z, the 
positive sense being that in which x increases? 



SBC. 9.1) PRODUCTS OF VECTORS 247 

Positive rotations. 

Before defining the vector product, we shall introduce a 
convention concerning the ^^ 

sign of a rotation. 

A rotation about a directed 

line L is said to be positive if 

it bears to the direction of L 
the same relation as the rota- 
tion of a right-handed screw ^ 
bears to its direction of travel FlG ' 100 - A positive rotation ' 
(Fig. 100). Thus a rotation from south to cast is a positive 
rotation about the upward vertical; the earth's rotation about 
its axis drawn from south to north is also positive. 

Right-handed triads. 

Consider three non-coplanar vectors. These three vectors, 
taken in some order, form an ordered triad. Since all the triads 
of which we shall speak are ordered, the adjective will be under- 
stood in future, and a triad will mean an ordered triad. Let 
P, Q, R be an orthogonal triad, the order being as indicated. 
This triad is said to be right-handed if the rotation through a 
right angle from P to Q is a positive rotation about R. Any 
other triad is said to be right-handed if it can be deformed con- 
tinuously into a right-handed orthogonal triad without its 
vectors becoming coplanar at any stage in the deformation. 

If the triad P, Q, R is right-handed, then the triad Q, P, R 
is said to be left-handed. 

If the triad of unit coordinate vectors i, j, k, introduced in 
Sec. 1.3, is a right-handed triad, the axes Oxyz are said to be 
right-handed. We shall always use right-handed axes for the 
sake of consistency. 

Vector product. 

Given two vectors P and Q, we draw the unit vector n per- 
pendicular to both P and Q, such that the triad P, Q, n is a 
right-handed triad. We define the vector product of P and Q, 
written P X Q, by 

(9.103) P X Q =PQsin0n, 

where is the angle between P and Q (Fig. 101). 



248 MECHANICS IN SPACE [Sue. 9.1 

It is clear from the definition that a rotation from P to Q, 
through an angle less than two right angles, is a positive rotation 

about P X Q. We note that the 
magnitude of P X Q is PQ sin 6} this 
is equal to the area of the parallelo- 
gram whose adjacent sides are P and 
PxQ Q. 

The vector product P X Q of two 
non-zero vectors vanishes if, and only 
if, P and Q are codirectional or oppo- 
site; in particular, 



Q P X P = 0. 

FIQ. 101. The vector product. T , / i j.i_ , 

Let us now find the components 

of P X Q. If we denote this vector by R, then R is perpendic- 
ular to both P and Q, and we have 



Therefore, 

(9.104) R z = fc(P 3 Qi - PiQ 3 ), 

- P 2 Qi), 

where k is an undetermined factor. Now, 
R* = fij + R\ + Rl 




= kWl + PI+ PD(Ql + Ql + Ql) 

- (Pid + P 2 Q 2 + P 3 0s) 2 ] 
= A; 2 P 2 Q 2 sin 2 0. 

But, by definition, 

R = PQ sin 0, 
and hence 

k = 1. 

From considerations of continuity, it is evident that k is to 
have the same sign in all cases. This sign may therefore be 
determined by considering the particular case where P and Q 
are unit vectors directed along the positive axes of x and y, 
respectively; then, 

P! = 1, P 2 = 0, P 8 = 0, 
Qi = 0, Q 2 = 1, Q 3 = 0, 



SBC. 9.1] PRODUCTS OF VECTORS 249 

and hence, by (9.104), 

Ri = 0, # 2 = 0, # 8 = k. 

But P X Q is, in this case, a unit vector directed along the positive 
axis of z, so that R s = 1. Hence, k = +1 here, and so in all 
cases. Thus, quite generally, the components of R = P X Q are 



(9.105) 



Ri = P 2 Q 3 - P 3 Q 2 , 
R* = P 3 Qi - PiQ 3 , 
R 3 = PiQ, - P 2 Qi. 



Note that the number describing the component and the sub- 
scripts in the leading term of the expression for that component 
are a cyclic permutation of the numbers 1, 2, 3. 

From the definition (9.103), it is evident that 

(9.106) P X Q = -Q X P. 
Again, using (9.105), it is easily shown that 

(9.107) PX(Q+R)=PXQ+PXR. 

Thus, vector multiplication is not commutative but does obey 
the usual distributive law for multiplication. 

For the unit coordinate vectors i, j, k, it is easily seen that the 
following relations hold : 

(9.108) 

1 j X k = i, k X i = j, i X j 

If we assume the distributive law for scalar and vector products 
and the formulas (9.108), we can establish (9.102) and (9.105) 
directly. Thus, 

P . Q = (Pii + P 2 j + P 3 k) (Qa + Q 2 j + Q 3 k) 



and 

P X Q = (Pii + P 2 j + P 3 k) X (Qii + Q 2 j + <?*k) 

= (P 2 Q 3 - P 3 Q 2 )i + (P 3 Qi - PiQ 3 )j + (PiQ 3 - P 2 Qi)k. 



If we multiply a vector by a scalar m, we do not alter its 
line of action; we merely change its magnitude, and reverse its 



250 MECHANICS IN SPACE [Sue. 9.2 

direction if m is negative. From this fact and the definitions 
of the scala,r and vector products, we see that 

(i) (mP) -Q = P- (wQ) - m(P. Q); 

(ii) (mP) x Q = P X (mQ) = m(P X Q). 

Hence, if a scalar factor appears in a product of vectors, 
its position is actually of no importance; it may b# shifted 
to any position without altering the value of the product as a 
whole. 

Exercise. If P X Q R and P X R Q, then the vectors Q and R 
both vanish. 

Differentiation of products of vectors. 

The derivative of a vector with respect to a scalar has been 
defined in Sec. 1.3. We saw there that the derivative of the 
sum of two vectors is equal to the sum of their derivatives, as 
in ordinary calculus. The ordinary rule holds also for the 
derivatives of the scalar and vector products. This is shown as 
follows: 

(9.109) / (P. Q) = lim (P + AP).(Q+AQ)-P.Q 
v ' du ^ ^' AU->O Aw 

r AP * Q + P * AQ + AP AQ 

"~ 



= ... 

du ^ du 

Similarly, writing "cross" for "dot," we obtain 

(9.110) jjL(PxQ)- XQ+Pxg- 

It is important to preserve the order of P and Q in (9.110), 
but not in (9.109). 

9.2. TRIPLE PRODUCTS 
Mixed triple product. 

Let us consider three vectors P, Q, and R. Prom them we can 
form the product P (Q X R), called their mixed triple product. 
This is the scalar product of P and the vector V = Q X R, and 
so is a scalar. We shall now express it in terms of the com- 



SEC. 9.2] 



PRODUCTS OF VECTORS 



251 



ponents of the three vectors. From (9.102) and (9.105), we 
have 

P (Q X R) = P - V 



or, in determinantal form, 



(9.201) 



p.(Q XR) = 



Pi Qi 
\Q\ 



From the rule governing the interchange of columns in a 
determinant, it follows that 

P (Q X R) = Q (R X P) = R (P X Q), 
and 

P (Q X R) = -P (R X Q) = -Q (P X R). 

Thus a mixed triple product is not changed by a cyclic permuta- 
tion of the vectors; its sign is reversed when two of the vectors 
are interchanged. 

We may interpret the mixed triple product geometrically as 
follows: As we have seen, the magnitude of Q X R is equal to the 
area of the parallelogram whose 
adjacent sides represent Q andR. 
Now P (Q X R) is the product 
of the magnitude of Q X R by 
the component of P in the direc- 
tion of Q X R (that is, per- 
pendicular to the plane of Q and 
R). Hence the magnitude of 
P . (Q X R) is equal to the vol- 
ume of the parallelepiped whose 
adjacent edges represent P, Q, and R (Fig. 102). The sign 
of P(Q X R) is also significant; it is positive or negative 
according as the angle between P and Q X R is acute or obtuse, 
i.e., according as P, Q, R form a right- or left-handed triad. 

From the geometrical interpretation, it is obvious that 

P-(Q XR) =0 
if the vectors P, Q, R are coplanar. 



Q x K 




Fjo 102 ._ ixed triple product . 



252 MECHANICS IN SPACE [Stec. 9.3 

Vector triple product. 

From the vectors P, Q, R, we can form another product, 
namely, P X (Q X R) ; this is evidently a vector and is called 
the vector triple product. 

We shall now express this product as the difference of two 
vectors. Denoting it by U and writing V = Q X R, we have 

U = P X V; 
hence, using (9.105), 
C/i = P 2 7 3 - P 3 F 2 



-(P< 

Similarly, 

17, = (P.R)Q S - 
17, = (P.R)Q- 

These three expressions for /i, C7 2 , C7 3 can be combined into the 
vector equation 

(9.202) U = P X (Q X R) = (P R)Q - (P Q)R. 

The following remark is an aid in remembering this expres- 
sion: since Q X R is perpendicular to the plane of Q and R, the 
vector P X (Q X R) must bo in this plane; hence, 

P X (Q X R) = qQ + rR, 
where q and r are scalars. 

Exercise. Evaluate all the vector triple products of the unit coordinate 
vectors i, j, k, including those in which one of the vectors is repeated. 

9.3. MOMENTS OF VECTORS 

The moment of a vector about a line was defined in Sec. 2.3 
as a scalar. There we spoke also of "the moment of a vector 
about a point A," but only as an abbreviation for "the moment 
about a line through A perpendicular to the plane containing A 
and the vector." Now that we are in possession of the powerful 
vector notation, we shall make a fresh start. We shall define 
the vector moment of a vector about a point, and (in terms of it) 



SEC. 9.3] 



PRODUCTS OF VECTORS 



253 



the scalar moment of a vector about a line; this latter definition 
will be shown to agree with that given in Sec. 2.3. 

Moment of a vector about a point. 

Let P be a vector with origin at B, and A any point in space 
(Fig. 103). We define the vector moment of P about A (or briefly 
the moment of P about A) as a vector M, given by 

(9.301) M = r X P, 

where r = AB, the position vector of B relative to A. Thus M 

D C 



1 


k M !&-' 


GX""^ 






\ 


k 




\ 


A 




\ j 




*X/a 


VX^A \ 


* X^ ] 


i^A P 


X^ \ 

? r 


/ 

i 



B 

FIG. 103. The moment of vector 
about a point. f 



FIG. 104. The moment of P 
about A is required. 



is a vector perpendicular to the plane of r and P, with magnitude 
(9.302) M = rP sin B = aP, 

where is the angle between r and P, and a the perpendicular 
from A on the line of action of P. 

As an illustration, let us calculate the moment of a given force P about a 
point A. In Fig. 104, P is a force of known magnitude applied at H and 
acting along the diagonal HF of one face of the cube ABC H. If i, j, k 
is a triad of unit orthogonal vectors at A (as shown), we have 



AH - 6(i + k), 
P - -^ ( J ~ ] 



where 6 denotes an edge of the cube. The moment M of P about A is now 
easily calculated; it is 

M - 6(1 + k) X ^= (j - k). 



254 MECHANICS IN SPACE [SBC. 9.3 

Hence, by (9.108), 



Thus the moment of P about A is a vector with components ( b 
6P/\/2 WVV2) in the directions of the edges AE t AB, AD, respectively. 

Returning to the situation shown in Fig. 103, let us investigate 
the effect of sliding P along its line of action. It "becomes 




B P B' P 

Fio. 105. The moment of a vector is unchanged when we slide the vector along 
its line of action. 

(Fig. 105) a vector P at B', where AB' = r + kP (k being some 
scalar). The moment about A is now 

M' = (r + fcP) X P. 
But P X P = 0, and hence 

M' = r X P = M. 

Thus the moment of a vector about a point is unaltered by sliding 
the vector along its line of action. 

We shall now make an important deduction from the above 
fact. Let P at B and P at B 1 be two vectors with a common 
line of action L, and let A be any point. Sliding P along 
L until its origin is at B, we do not alter its moment about A. 

Thus, if AB = r, the sum of the moments about A of P at B 
and -Pat B' is 

r X P + r X (-P) = r X (P - P) = 0. 

In words, for two vectors in the same line, with equal magnitudes 
but opposite senses, the vector sum of moments about any point 



SBC. 9.3) PRODUCTS OF VECTORS 255 

is zero. In particular, by the fundamental law of action and 
reaction (cf. Sec. 1.4), we have 

(9.303) The vector sum of moments about an arbitrary point of 
the forces of interaction between two particles of any system is zero. 

By the distributive law for vector multiplication, we have, 
for any vectors, 

(9.304) rXP + rXQ+rXR+-- 



Let P, Q, R, be vectors with common origin B, and let r 
be the position vector of B relative to a point A. Then, for 
vector moments about a point, we have the theorem of Varignon 
(cf . Sec. 2.3) : The sum of the vector moments about a point A of 
vectors P, Q, R, with common origin B, is equal to the vector 
moment about A of the single vector P + Q + R + with origin 
B. 

Moment of a vector about a line. 

Let M be the moment of a vector P about a point A, and 
let L be any line through A. Of the two senses on L, we choose 
one as positive and distinguish it by 
a unit vector 3. lying on L. We 
define the scalar moment ofP about L as 
the component M\ of M in the posi- 
tive sense of I/; expressed in symbols, 

(9.305) MX = 3t.M. 

We shall now show that the above > 

definition is equivalent to that given 
in Sec. 2.3. We take special axes 
Oxyz as shown in Fig. 106; the origin 
coincides with A, and Oz lies along 
the line L in the positive sense. 
Relative to these axes, P has components (X, F, Z) and acts at a 
point B with coordinates (x, y, 2). The moment M of P about A 
(or 0) is 

(9.306) M - (xi + y] + *k) X (XI + Yj + Zk) 

= (yZ - zY)l + (zX - zZ)j + (xY - 




/ 

*R 



256 MECHANICS IN SPACE [Sac. 9.3 

where i, j, k are the unit coordinate vectors. Since & = k, 
we have 

(9.307) MX = k M = xY - yX. 

But this quantity is precisely the moment as given by (2.303); 
the two definitions of the moment of a vector about a line are 
now completely reconciled. 

It might appear that the value of the moment MX of P about L, 
as given by (9.305), depends on the choice of a point A on this 
line. This is not actually the case. For let A' be any other 

point on L, so that A A' fa, where k is some scalar. The 
moment about A' of P at B is 

M' = (-ASt + r) XP, 

where r = AB. The component of M' in the positive sense of 
L is therefore 

M( =3i'[(-Wi + r) XP] 

= -A&.(3L XP) +3i-(rXP) 
= 31 M = MX, 

since A (3i X P) =0. 

The theorem of Varignon for scalar moments of vectors about 
a line follows directly from (9.304) ; we have merely to take the 
scalar product of each side with X This very simple proof by 
vector methods should be compared with that of Sec. 2.3, where 
only elementary methods were used. 

There are occasions, however, where scalar methods are more 
direct than vector methods. On such occasions, we require 
formulas for the moments of a vector about the axes of coordi- 
nates. These are, by (9.306), 

(9.308) yZ - zY, zX - xZ, xY - yX, 

where (X, Y, Z) are the components of the vector applied at 

(z,2/, z). 

Exercise. A vector with components (1, 2, 3) acts at the point (3, 2, 1). 
What is its moment about the origin, and what are its moments about the 
coordinate axes? 



SBC. 9.4] PRODUCTS OF VECTORS 257 

9.4. SUMMARY OF PRODUCTS OF VECTORS 

I. Scalar product. 

(9.401) P Q = Q P - PQ cos B = PjQj + P 2 Q 2 + P 8 Q 3 . 

II. Vector product. 

(9.402) P X Q = -Q X P = PQ sin 0n. 

(n a unit vector perpendicular to P and Q; triad P, Q, n right- 
handed.) 

III. Usual rules of algebra and calculus apply to products of 
vectors, if order in vector products is preserved. 

IV. The mixed triple product. 

Pi On 

(9.403) P.(Q XR) = P 2 Q 2 J 

Pa Qa J 

V. The vector triple product. 

(9.404) P X (Q X R) = (P - R)Q - (P Q)R. 

VI. Moment of a vector about a point. 

(9.405) M = r X P 

= (yZ - zY)i + (zX - rcZ)j + (xY - 

VII. Moment of a vector about a directed line O). 

(9.406) MX = * (r X P). 

EXERCISES IX 

1. Solve the equations 

2A + B = M, A + 2B - N, 

M and N being given vectors. 

2. Three vectors arc represented by the diagonals of three adjacent 
faces of a cube, all passing through the same corner and directed away from 
it. Find their sum. 

3. What is the moment about the re-axis of a force of magnitude 3 applied 
at a point with coordinates (2, 3, 5), in a direction making angles of 60 
with the axes of y and z and an acute angle with the axis of x? 

4. A, B, C, D are any four vectors. Prove that there exist scalars a, 6, 
c, d (not all zero), such that 

oA + &B + cC + dD = 0. 



258 MECHANICS IN SPACE [Ex. IX 

5. If A X B A X C, show that B C -H &A, where k is some scalar. 

6. If A and B are any two unit vectors, prove that the moment of A 
about B is equal to the moment of B about A. 

7. Find the moments, about a corner of a cube, of three unit vectors 
converging on the opposite corner along three edges. Show that the sum 
of the moments is zero. How could you obtain this result without 
calculation? 

8. A force with components (X, Y, Z) acts at the point (a, 6, c). What 
is its moment about a line through the origin with direction cosines (I, m, n) ? 

9. A force of magnitude P acts along the line joining opposite corners 
of a cube of edge 2a. Find the moment of the force about a line which is 
a diagonal of a face of the cube and which does not cut the line of action 
of the force. 

10. A directed line L passes through the point (a, 6, c) with direction 
cosines (I, m, n). Prove that the moment about L of a unit vector pointing 
along the z-axis is bn cm. 

11. Prove that the moment of a vector about a line vanishes if, and only 
if, the vector cuts the line or is parallel to it. 

12. Prove the identities 

(i) A X (B X C) + B X (C X A) + C X (A X B) - 0, 

(ii) A X [B X (C X D)J = (B D)(A X C) - (B C)(A X D), 

(iii) (A X B) X (C X D) = B[A (C X D)] - A[B (C X D)]. 

13. Oxyzj Ox'y'z' are two sets of rectangular Cartesian axes. P is a vector 
with components X, Y, Z on Oxyz and components X', Y' t Z' on Ox'y'z 1 . 
Show that 

X' - a n X + ai Z Y + On Z, 

where an, ais, AH are the direction cosines of Ox' with respect to Oxyz. 
Develop similar formulas for Y' and Z'. 

14. Solve the differential equation 



where a is a constant vector. 

15. Show that the differential equation 



where a and b are perpendicular constant vectors, has the general solution 

r-/(fla + fc + e - gj (a X b); 
here/(i) is an arbitrary function and c, e are arbitrary constant vectors. 



CHAPTER X 
STATICS IN SPACE 

10.1. GENERAL FORCE SYSTEMS 

Before proceeding to conditions of equilibrium, let us develop 
some results valid for any system of forces, whether they produce 
equilibrium or not. 

The total force and the total moment. 

Let there be a system of particles with position vectors TI, 
r 2 , T n relative to a point 0, and let forces PI, P 2 , P n 
act on them. We define the total force F of this system as the 
vector sum of the forces, i.e., 

(10.101) F = 2) P.. 

The moment of the force P about O is r, X P, by (9.301). 
We define the total moment G of the force system about the 
base point as the sum of these moments, i.e., 

n 

(10.102) G = ] r f X P*. 

The scalar components of the total force and the total moment 
on axes Oxyz are easy to write down. Let xi, y t , z l be the coordi- 
nates of the tth particle and X it Y iy Zi the components of P t . 
Then the components of F are 

(10.103) x = 5) x t , Y = 5) Y if z = 2) z t , 

and the components of G are 
(10.104) L - 

N 
259 



260 MECHANICS IN SPACE [Ssc. 10.1 

Since the scalar moment about Ox is the component along Ox of 
the vector moment about 0, it is evident that L, M, N are the 
total scalar moments about the axes Oxyz ; thus, for example, L 
is the sum of the scalar moments of all the forces about Ox. 

Change of base point. 

It is clear that F does not depend on the choice of base point 0. 
On the other hand, G does depend on this choice. Let us see 
how G changes when we change the base point from to 0', 

where 00' = a. 

If rj, ig, r' n are the position vectors of the particles 
relative to 0', we have 

(10.105) Ti = r( + a. 

Then, if G' is the total moment about 0', we have 

G' = ri X P, 

t=l 

= J) (r, - a) X P,; 

1 = 1 

and so, by (10.101) and (10.102), 

(10.106) G' = G - a X F. 

This equation shows how the total moment changes with change 
of base point. 

Equipollent force systems. 

In Sec. 2.3, we gave the general definition of equipollence 
but used it only in the restricted sense of plane equipollence. 
We recall that two force systems are equipollent if (in the lan- 
guage used above) the total forces of the two systems are equal, 
and also their total scalar moments about an arbitrary line. We 
shall now establish the following fundamental theorem: 

// two force systems are equipollent, they have the same total 
force and the same total moment about an arbitrary base point 
the same for both systems. Conversely, if two force systems have 
the same total force and the same total moment about some one base 
point t then they are equipollent. 



SBC. 10.2] STATICS IN SPACE 261 

Let Si and S 2 be two force systems and any base point. 
The total forces of the two systems will be denoted by FI, F 2 , 
and their total moments about by Gi, G 2 , respectively. 

If /Si and & are equipollent, then FI = F 2 from the definition 
of equipollence. Further, the scalar moments of Si and S 2 
about any line through are equal, and so the vectors GI and G 2 
have the same component along any line through 0; hence 
Gi = G 2 . Thus, for an arbitrary base point 0, we have 

(10.107) F! = F 2 , G! = G 2 , 

which establishes the first part of the theorem. 

To prove the converse, we must show that Si and Sz are 
equipollent if (10.107) hold for some one base point O. The 
first condition of equipollence, namely, the equality of total 
forces, is evidently satisfied; it remains to prove that the scalar 
moments of Si and $ 2 about any line are equal. If the line 
passes through 0, the equality of scalar moments follows at 
once by projecting the equal vectors GI, G 2 on the line. If the 
line does not pass through 0, let 0' be- any point on it. The total 
moment GJ of Si about 0' is expressed in terms of FI and Gi 
as in (10.106); there is a similar expression for the total moment 
G 2 of Sz about 0'. Those vectors are obviously equal by virtue 
of (10.107), and so the scalar moments in question are also equal. 
The proof of the theorem is now complete. 

If the total force F of a system is zero, and also the total 
moment G about some one. base point, it follows that F and G 
are zero for all base points. We say then that the system is 
equipollent to zero. 

10.2. EQUILIBRIUM OF A SYSTEM OF PARTICLES 
In Chaps. II and V, we developed the general principles 
of statics and dynamics in a plane. In establishing these 
principles, we sometimes gave the results in three-dimensional 
form, where there was no particular difficulty involved. Now 
we have to develop general principles in three dimensions, 
and it might be thought that the new work would have to be 
built on top of the old. That is not the case. Since we are 
now in possession of the powerful vector method, it is on the 
whole simpler to establish the general principles directly from 
the basic laws of Sec. 1.4. That is what we shall do, except 



262 MECHANICS IN SPACE (SEC. 10.2 

in those cases where the vector method offers no advantage. 
In the main, therefore, the rest of the book is logically inde- 
pendent of Part I, except for the laws of Sec. 1.4. 

Necessary conditions of equilibrium. 
For a single particle, the condition of equilibrium is 

(10.201) P = 0, 

where P is the vector sum of the forces acting on the particle. 

Let us now consider a system of n particles in equilibrium. 
Let P denote the resultant of the external forces acting on the 
ith particle. In addition to the external forces, there act on 
each particle a number of internal forces due to the other particles 
of the system. Let P# denote the force on the ith particle due 
to the jth particle; by the law of action and reaction, these inter- 
nal forces satisfy 

(10.202) P t/ + P, t = 0. 

Since each particle is in equilibrium, it follows from (10.201) 
that the external and internal forces satisfy the equations 

P! + 04- Pi2 + Pis + ' + Pm = 0, 

(10.203) p 2 + Psi + + P 23 + + P 2 n = 0, 

Pn + Pnl + Pn2 + ' ' ' + P,n-l + = 0. 

When we add these equations, the internal reactions cancel on 
account of (10.202), and so 

(10.204) F = 0, 

where F is the total force of the external force system, viz., ] P- 



Let r denote the position vector of the ith particle relative 
to a base point 0. If we multiply the equations (10.203) vec- 
torially by ti, r 2 , r n in order, we obtain 

X Pi + + ri X Pi2 + n X Pis + 

+ ri X Pm = 0, 

r 2 X P 2 + r 2 X P + + r 2 X P 2 s + 
(10.205) + r 2 X P 2 n - 0, 



r n X P + r X Pi + r X P2 + 

+ r n X P n ,-i + = 0. 



SBC. 10.2J STATICS IN SPACE 263 

Now, 

ri X Pi2 + r 2 X P = (ri - r 2 ) X Pi 2 0, 

since ri r 2 and Pi 2 lie in the same line [cf. (9.303)]. Hence, 
on addition of the equations (10.205), the terms symmetrically 
placed with respect to the line of zeros cancel in pairs, and we 
get 

(10.206) G = 0, 

where G is the total moment about of the external force 

n 

system, viz., J) r X P.. 



Thus, for a system in equilibrium, F and G both vanish, and so 
(in the language of Sec. 10. 1) we have the following general result : * 
// a system of particles is in equilibrium, then the external force 
system is equipollent to zero. 

In terms of the total force F and the total moment G about 
any base point 0, the above statement is equivalent to the two 
vector equations 

(10.207) F = 0, G = 0. 

Resolving vectors along rectangular axes Oxyz and using the 
notation of (10.103) and (10.104), we get the following six scalar 
conditions of equilibrium: 

(10.208) X = 0, Y = 0, Z = 0; 

(10.209) L = 0, M = 0, N = 0. 

In this form the conditions appear as generalizations of (2.308). 
The whole theory of statics rests on the equations (10.207). 
Most frequently, these equations are applied to a rigid body, 
treated as a whole. But they are valid for any system, which 
may be a part of a rigid body or a piece of an elastic material, or 
even a volume of fluid. As an example, we shall presently 
discuss the equilibrium of a flexible cable in space (cf. Sec. 3.4 
for the plane case). The equilibrium of a rigid body will be 
considered in more detail in Sec. 10.4. 

* This condition is equivalent to the conditions obtained in Sec. 2.3. 



264 MECHANICS IN SPACE [SBC. 10.2 

Curves in space. 

We require some elements of the geometry of curves in space; 
they are of importance apart from the present connection and 
will be used again later. 

Let C be a curve in space and A any point on C. Let P be any 
other point on C, distant s from A (s being measured along the 
curve). The unit vector i, tangent to the curve at P, is clearly 
a vector function of s. Since i i remains equal to unity along 
C, we have 

(10.210) i'Ts"' 

It follows that the vector di/ds is normal to C at each point P. 
Let 1/p (p is the radius of curvature of C at P) denote the magni- 
tude of this vector. Then we may write 

(10.21D I - i, 

where j is a unit vector normal to C; it is the unit principal 
normal vector. The plane of i and j is called the osculating plane. 

The unit binormal vector k at P is defined as follows: It is 
normal to both i and j and is so directed that (i, j, k) is a right- 
handed triad. 

The equation (10.211) is the first of the Frenet-Serret formulas. 
The complete set of formulas is 

di j dj k i dk_ j 

-T" > ~7~ = ~~ > T~ == ' 

as p ~ ds T p ds T 

where r is a certain scalar, called the radius of torsion. * These 
formulas are easily proved. Since 

j.-0, k.-0, 

J ds ' ds ' 

it follows that 

(10.213) g = ai + 6k, * = ai + ffl, 

* We note that p is necessarily positive, since it is defined as the reciprocal 
of the magnitude of di/ds} r may be positive or negative. 



Sac. 10.21 STATICS IN SPACE 265 

where a, 6, a, ft are scalars. On differentiating the relations 

(10.214) i-j=0, j-k = 0, k-i-0 
and using (10.211) and (10.213), we find 

(10.215) a = - ^ a = 0, 6 + = 0. 



Hence, writing b = 1/r, we obtain (10.212). 

For a curve C, drawn on a surface , the vector i is necessarily 
a tangent to S. But the vector j is not necessarily normal to 
S; it may even be a tangent to S, as in the case where S is a 
plane. If j is normal to S at each point of C, the curve is called 
a geodesic on S. 

Flexible cables. 

Let us now consider a flexible cable in equilibrium under the 
action of known external forces and the tensions at its ends. 






-Ti 



k 

FIG. 107. Forces on an clement of cable. 

Figure 107 shows an infinitesimal portion PQ, P being at a 
distance s from one end of the cable. Let i, j, k denote the unit 
tangent, principal normal, and binormal vectors at P. The 
forces acting on the element PQ (length ds) may now be described 
as follows: 

(i) a force Ti at P, where T is the tension at P; 

(ii) a force (T + dT)(i + di) = (T + dT)[i + (j/p) da] at Q, 
where T + dT is the tension at Q; 

(iii) a force R ds (Rii + R%j + 72 gk) ds, where R is the 
external force per unit length of the cable. (This force acts at 
some unspecified point of the element PQ.) 

The element PQ is a system in equilibrium under these forces, 
and so we may apply the conditions (10.207) to it. From the 
first of these conditions, we have 




266 MECHANICS IN SPACE [Sao. 10.3 

(10.216) - Ti + (T + dT) (i + i da) 

+ Rtf + flak) da = 0. 



The second of the conditions (10.207) is satisfied identically to the 
first order in ds. From (10.216), we at once obtain the scalar 
equations 



(10.217) 



R* = 0. 



These are the general equations of equilibrium. They enable us 
to find the form of the cable and also the variation in tension 
along it. In particular, the last of these equations tells us that 
the osculating plane at each point contains the external force 
vector. 

Example. A light cable rests in contact with a smooth surface S, under no 
forces except the reaction of S and the tensions at its ends. It is required to 
find the curve C in which the cable rests t and also the tension at each point. 

The external force vector R ds is the reaction of the surface S on the 
element. Since this reaction is normal to S, it is also normal to C, and so 
Ri - 0. But fl s 0, by the last of (10.217), and so 

R - # a j. 

It follows that the principal normal vector j is normal to S at each point of C. 
Hence, C is a geodesic on S. Thus, to construct a geodesic joining two given 
points on a surface, we have merely to stretch a light thread between these 
points. If S is a sphere, C is an arc of a great circle; if S is a cylinder, C is a 
curve on the cylinder which maps into a straight line, when the cylinder is 
cut along a generator and unrolled on a plane. 

Again, since R\ 0, the first equation in (10.217) gives 

- 

Thus the tension T is constant; in particular, the tensions at the ends are 
equal. 

10.3. REDUCTION OF FORCE SYSTEMS 

If we succeed in finding a simple force system S f , equipollent 
to a given system S, we say that we have reduced the system S 
to the system S'. We shall presently consider, in some detail, the 



SEC. 10.3] 



STATICS IN SPACE 



267 



reduction of force systems; but, before doing this, it is convenient 
to have a vector description of the particular force system known 
as a couple. 

Moment of a couple. 

As in Sec. 2.3, a couple is defined as a pair of parallel forces, 
equal in magnitude but opposite in sense. Figure 108 shows a 
couple consisting of the forces P and P applied at the points 
A and B, respectively; is any point in space. The vector 



-P 




G=pxP 

FIG. 108. The moment of a couple. 



moment G of this couple about is easily found; denoting OB by 
r and BA by p, we obtain 

(10.301) G = r X (-P) + (r + p) X P = p X P. 

This value of G is independent of the position of the point 0. 
In other words, a couple has the same moment about all points 
in space. Thus the vector G may be regarded as a free vector; 
it is perpendicular to the plane determined by the forces P, P 
of the couple; its magnitude is p'P, where p' is the perpendicular 
distance between the lines of action of these forces (Fig. 108). 

Since two couples which have the same moment are equipollent 
force systems, a couple is completely specified (as far as equi- 
pollence is concerned) by its free moment vector, or, briefly, its 
moment. Thus, when we speak of a couple G, we have in mind 
any one of an infinite number of couples, each of which has 
moment G. 

To avoid confusion in diagrams, the arrowheads indicating 
couples may be marked with a crossbar, as in the figure. 



268 MECHANICS IN SPACE [Sue. 10.3 

Composition of couples. 

Let there be a system of forces consisting of a number of 
couples Gi, G 2 , . The total force of this system is zero, and 
the total moment about any point is clearly 

(10.302) G = Gi + G 2 + - . 

Thus, a system consisting of couples is equipollent to a single 
couple; its moment is equal to the vector sum of the moments of the 
individual couples. In other words, couples are compounded 
by the parallelogram law. 

Exercise. Forces with components (2, 0, 0), ( 1, 0, 0), ( 1, 0, 0) act at 
the points (0, 0, 0), (0, 1, 0), (0, 0, 1), respectively. Show that they can be 
reduced to a couple, and find its magnitude and direction. 

Reduction of a force system to a force and a couple. 

Consider a general force system S, with total force F and total 
moment G with respect to a base point 0. Consider also a 
Q second force system 8', consisting only of a 

F single force F applied at O and a single couple 
G. Obviously, 8' is equipollent to S. There- 
fore a general force system can always be reduced 
to a single force applied at an arbitrary base 
point, together with a couple. 
O Just as we represent a single force by an 

FIG. 109. Repre- arrow, so we can represent a general force 

sentation of a gen- , , ,. , ji . TV i/w\ 

eral force system by system by a diagram such as that in *ig. 109; 
a force F and a this shows a force F acting at a base point 
coupe ' and a couple G. Although the couple G is 

a free vector, it is convenient to draw it out from the base 

point 0. 

> 
If we change the base point from to 0', where 00' = r, we do 

not alter F, but the moment about 0' is not G; it is found by 
adding to G the moment of F about 0'; this gives [cf. (10.106)] 

G' = G - r X F. 

Hence, under a change of base point from to 0', the force F 
and the couple G become F' and G', respectively, where 

(10.303) F' = F, G' = G - r X F. 




SBC. 10.3] STATICS IN SPACE 269 

We note that F' = F and F' G' = F G; in words, the scaJars 
F and F G are invariant under a change of base point. If either 
of these invariants vanishes for one choice of base point, then it 
vanishes for all choices of base point. 

Reduction to a wrench. 

A wrench consists of a force F and a couple G with parallel 
representative line segments. This relation between F and G is 
expressed by the vector equation 

(10.304) G = pF, 

where p is some scalar having the dimensions of a length. The 
quantities p and F arc called the pitch and intensity of the wrench, 
respectively. The line of action of the force F is called the axis 
of the wrench. 

A general force system can always be reduced to a wrench. 
We shall now show how this is done. Let us first reduce the 
system in question to a force F at a base point and a couple G. 

Changing the base point to 0', where 00' = r, we obtain a force 
F' and couple G'. These constitute a wrench if 

(10.305) G' = pF'. 
Since, by (10.303), 

F' = F, G' - G - r X F, 

the equation (10.305) is satisfied if r and p satisfy 

(10.306) G - r X F = pF. 

This is, in fact, a vector equation for r (the position vector of 0') 
and p (the pitch of the wrench). 

Let us take as origin of rectangular Cartesian coordinates, 
and let (Fi, F 2y FZ), (Gi, G%, (73) denote the components of F, G, 
respectively. The vector equation (10.306) is equivalent to the 
scalar equations 

gi ~ yFt + zFz _ G 2 - zFi + xF 9 _ OB - xF z + yF l _ 

7[ Fl F, - P > 

where x, y, z are the coordinates of 0'. These equations show 
that F', G' constitute a wrench provided 0' lies on the straight 



270 MECHANICS IN SPACE [SEC. 10.3 

line with equations 

Gl ~ V F * + zF * G * ~ zFl + xF * G * ~ xF * + ^ Fl 



- - - ft - - p -- 

r 1 r 2 T3 

We note that, if (x, y, z) is any point on this line, then (x + kFi, 
y + kFz, z + kF s ), where k is any scalar factor, is also on it. It 
follows that this line has the direction of F; it is the axis of the 
wrench to which the system is reduced. 

The pitch p is found by taking the scalar product of F and the 
vectors on the two sides of (10.306). We find 

F-G = pF 2 ; 
therefore, 



(10.308) p = 



It is, of course, not accidental that the pitch of the resulting 
wrench is a function of the invariants F and F G. 

If p = 0, the wrench degenerates into a single force; in this 
case F G = 0. If p is infinite, the wrench degenerates into a 
couple; in this case F = 0. In each of these special cases, we 
have a force system equipollent to a plane system of forces. 
Conversely, if the force system is equipollent to a plane system 
of forces, then one or other of these special cases must arise. 

Exercise. In the reduction of a given force system to a force and a couple, 
the couple G depends on the base point. For what base points is G least? 

Reduction of a system of parallel forces. 

A set of parallel forces is a system of particular importance 
in mechanics, e.g., the weights of a number of particles. 

Any ft parallel forces may be denoted by /bJP, & 2 P, k n P, 
where fci, & 2 , * k n are scalars. Selecting a base point 0, we 
first reduce this system to a force F at and a couple G. Let 
r, (s = 1, 2, n) denote the position vectors, relative to 0, of 
the points of application of the several forces. Then, 



(10.309) 



F - (W - fcP > 



( r - X *- p ) - (2 fc ' r ') X P = r X F, 



Sac. 10.3] STATICS IN SPACE 271 

where 

(10.310) k = ]g k,, r = ( V k,T.)/k. 

i \-i ' 

This reduced system is clearly equipollent to the single force F, 
applied at the point C with position vector r. 

Thus a system of parallel forces can be reduced to a single 
force, unless A; = 0. If k = 0, it can be reduced to a couple. 

The point C, with position vector r given by (10.310), is 
called the center of the system of parallel forces. Its position is 
determined solely by the vectors r 8 and the ratios of the numbers 
ki, k%, k n . It is unaltered by turning the several forces 
about their points of application, provided they retain their 
magnitudes and remain parallel to one another. 

If the forces in question are the weights of the particles of a 
system, the point C is the center of gravity of the system (cf. 
Sec. 3.1). 

The reduction of some special force systems. 

The force systems encountered in practical problems are often 
extremely complicated. However, the details of such a force 
system are relatively unimportant when it acts on a rigid body; 
if we know the total force and the total moment, we know all that 
is essential for the discussion of equilibrium. The total force 
and the total moment play an equally important part in dynamics 
(see Chap. XII). 

1. Analysis of forces on an airplane. 

Figure 110 shows an airplane in flight. C is the mass center. The 
orthogonal right-handed triad of unit vectors i, j, k is fixed in the airplane; 
j is perpendicular to the plane of symmetry and points to the right; i and k 
lie in the plane of symmetry. The direction of i is fixed in some conven- 
tional way (e.g., parallel to tho airscrew axes), so as to be nearly horizontal 
and point forward when the plane is in normal flight; k will then be directed 
nearly vertically downward. 

The forces acting on the airplane are as follows: 

(i) The weights of the various parts. These constitute a system of 
parallel forces and can be reduced to a single force W acting vertically down- 
ward through C\ W is the total weight of the airplane. 

(ii) The thrust, or driving force, P due to the airscrews. This force is in 
the direction of the vector i or nearly so. 

(iii) Forces arising from the action of the air. These forces are due 
mainly to variations in pressure over the wing surface and in a lesser degree 



272 MECHANICS IN SPACE [SEC. 10.3 

to frictiftn ; they form a very complicated system. Reducing this system to 
a force F at C and a couple G, we resolve as follows: 

F = Xi + Yj + Zk, G - Li + Mj + Nk. 

For normal flight, X is the drag, Z is the lift, and M is the pitching 
moment; Y t L, N are zero. The precise terminology of aerodynamic theory 
is not quite so simple as this. Under normal flight conditions, however, the 
differences are small. 

The theoretical determination of the force system (F, G) is a* very difficult 
problem in hydrodynamics; and, in practice, experimental methods are used. 
A model of the airplane (or the airplane itself) is mounted in a wind tunnel. 
Direct measurements are then made of the force system required to keep the 
model at rest in a stream of air. 




k 

FIG. 110. Reference vectors for an airplane. 

An analysis of the forces on a bullet or shell follows the same lines. In 
this case, however, the driving force P is absent. 

2. Analysis of stresses in a beam. 

Consider a beam in equilibrium and let Ox be a line in the direction of its 
length. We imagine the beam cut in two by a plane II perpendicular to Ox 
at A (Fig. 111). We shall denote by R the part of the beam to the right of n, 
and by L the part to the left. 

The forces on L are as follows: 

(i) Applied forces, such as gravity or external loads. These are equi- 
pollent to a single force F at A and a couple G. 

(ii) Forces exerted across n by R on L. These are internal forces for the 
whole beam, but external forces for the system L. They are called the 
stresses across the plane section n and are equipollent to a force S at A and a 
couple M. Introducing the orthogonal triad of unit vectors i, j, k, as shown, 
we write 

S * Sd + Sd + S 8 k, M - Mil + M 2 j 
The following terminology is used: 

Si =* tension, 

2, $3 = shearing forces, 

Mi twisting couple, 

M 2 , M 3 bending moments. 



SEC. 10.4] 



STATICS IN SPACE 



273 



If the applied forces are known, then F, G are known and we can find S, M. 
We have merely to apply the conditions of equilibrium (10.207) to L, 
obtaining 

S - -F, M - -G. 

If we change the section II by varying the distance x of A from 0, F and Gare 
known functions of x\ hence, S and M are known functions of x. In other 
words, there are two vector functions S(z), M(x) [or six scalar functions 
Si(x), S z (x), - M s (x)] which give, for each value of x, a force system 
equipollent to the stresses across the corresponding cross section of the beam. 
Failure in an engineering structure, such as a bridge, is due to excessive 
stress. The engineer must know in advance if any given beam or girder 
is likely to fail under the loads which it will be called on to support. 




FIG. 111. Reference vectors for reactions in a beam. 

Although the values of S and M do not give a complete picture of the 
internal stresses, they are the quantities which the engineer calculates 
in order to see whether or not a structure is safe. 

The above analysis also applies in naval architecture. Regarding the 
hull of a ship as a beam, subject to known applied forces, we can determine a 
force system (S, M) equipollent to the internal stresses across any section 
perpendicular to its length. A ship must be so constructed that it will with- 
stand the action of stresses (S, M) arising from the applied forces of weight 
and buoyancy. In a storm the ship may be supported by waves under bow 
and stern; then the forces of buoyancy are concentrated there, and the ship is 
in danger of "breaking its back." 

10.4. EQUILIBRIUM OF A RIGID BODY 
Necessary and sufficient conditions of equilibrium. 

In our mathematical model, a rigid body is a set of particles 
whose mutual distances are invariable. Let us now consider a 
rigid body acted on by external forces. Reducing the force 



274 



MECHANICS IN SPACE 



[SEC. 10.4 



system to a single force F at a base point and a couple G, we 
know by (10.207) that the conditions 



(10.401) 



F = 0, G = 



R 




are necessary for equilibrium. We now make use of the rigidity 
of the body to prove that these conditions are also sufficient, so 
that the body must be in equilibrium if they are satisfied. 

Let us suppose that a rigid body acted on by external forces, 
satisfying (10.401), is not in equilibrium; then the particles 

of the body will be on the point 
of moving. This motion will be 
prevented by introducing the 
following constraints (Fig. 112) : 
(i) The point (taken in the 
body) is fixed; this leaves the 
body free to turn about 0. 

(ii) With origin 0, we draw a 
unit vector i; the particle A at 
its extremity is constrained to 
slide in a smooth tube with axis 
in the direction of i. The two 
constraints now introduced fix 
all points of the body on the 
line OA, but still permit the body to turn about this line. 

(iii) With origin 0, we draw a unit vector j, perpendicular 
to i; the particle B at its extremity is constrained to move 
between two smooth planes parallel to the plane OAB. If these 
planes are close to each other, this constraint will prevent the 
motion of B. 

These three constraints together prevent any motion of the 
body, and so it must remain at rest. It is therefore in equi- 
librium under the action of the given external force system and 
the reactions of these constraints. 

Now the reactions of constraint are equipollent to a force 
F' at and a couple G'. Since the body is in equilibrium, 

F + F' - 0, G + G' = 0; 

and so, by (10.401), 

(10.402) F' - 0, G' - 0. 



FIG. 112. Constraints preventing mo- 
tion of a rigid body. 



SBC. 10.4] STATICS IN SPACE 275 

In view of the smoothness of the constraints at A and J5, we 
see that the forces of constraint are 

(i) a force P = PJ + P 2 j + Psk applied at 0; 

(ii) a force Q = Q 2 j + Qak applied at A (position vector i 

relative to 0) ; 

(Hi) a force R = RJs. applied at B (position vector j relative to 0) . 

The vector k is a unit vector completing the orthogonal triad 
i, j, k. On calculating F, G' in terms of Pi, P 2 , P 3 , Qz, Qs, Rz, we 
find that the six scalar equations contained in (10.402) imply 

Pi = P 2 = P 3 = Q 2 = Q 3 = #3 = 0. 

Hence the constraints introduced actually exert no reactions, and 
so the body remains in equilibrium even if they are removed. 
The sufficiency of the conditions (10.401) is now established. 

It is possible to state the conditions of equilibrium in forms 
other than (10.401). For example, it is easy to see that, if the 
external forces have no moment about each of three non-collinear 
points, then the conditions (10.401) are satisfied. Conversely, 
if the external forces satisfy (10.401) for some particular base 
point, they have no moment about any point. Thus, if G, G', G" 
denote the total moments of the external force system about each 
of three non-collinear points, the conditions 

(10.403) G = G' - G" - 

are both necessary and sufficient for equilibrium. Occasionally 
the conditions (10.403) are easier to apply than (10.401). 

Applications. 

The conditions (10.401) will now be applied to solve two problems. 

Example 1. Figure 113 shows a pulley, with radius r and center A, rigidly 
attached to a horizontal shaft BCD. This shaft is free to turn in smooth 
bearings at B and C; the end D projects beyond the bearing at C, and to it is 
rigidly fastened a crank DE with handle EH. The angles BDE and DEH 
are right angles. A weight W is attached to the lower end of a cord passing 
round the pulley, the other end of the cord being fixed to the pulley. To 
raise W, a man applies a force P at // in a direction perpendicular to BCD 
and making an angle <f> with the horizontal. 

It is required to find the magnitude of P and also the directions and magni- 
tudes of the reactions R and R' at B and C, respectively. 



276 



MECHANICS IN SPACE 



[SEC. 10.4 



Let i, j, k be an orthogonal triad of unit vectors at A, i lying along AC 
and k pointing vertically upward. Lengths are denoted as follows: 
BA = AC = a, CD 6, DE = c, EH = d. Then, if DE makes an angle 
with the vertical, the forces acting on the whole system can be described as 
follows (position vectors being taken relative to A): 



(10.404) 



a force P = P cos </> j P sin < k 
at (a + 6 -f 
a force Wk at rj, 
a force R = R 2 j + Rfc at ai, 
a force R ; = #Jj H- /e' 3 k at ai. 



, + c sin j + c cos 6 k, 




t-Wk 

FIG. 113. Pulley and shaft turned by a crank. 
Reducing this force system to a force F at A and a couple G, we find 



F = (P cos </ + R 2 + R' 2 )j + (-P sin <t> - W + # 3 + 
G = [(a + b + d)i + c sin j -f c cos k] X [P cos ^ j - P sin k] 
+ rj X Wk - ai X (fl J + # 3 k) + ai X (/& + 
= [JTr - PC cos (0 - *)]i + [P(a + *> + d) sin * + a,K 3 - afl' 8 ]j 

cos <*> - aR 2 



For equilibrium, these vectors must vanish. Equating them to zero and 
performing some simple calculations, we find 



(10.405) 






-- "~ P sm 



, Tf 
C 3 = -- +P 



sin *' 



These equations constitute the solution of our problem. 

The value of P given above is least when <f> = 0, i.e., when the force at // is 
perpendicular to the crank DE; this may also be seen quite simply by taking 
moments about the line BD. Thus, to raise the weight with the least effort, 
the man should push at right angles to the crank. As far as the man is con- 
cerned, the actual position of the point // in the handle is of no importance; 
a change in d merely alters the reactions at B and C. 



SBC. 10.7] 



STATICS IN SPACE 



297 



this is, therefore, the only position of equilibrium, 
observe that 



As for stability, we 




Since V = for qi q$ and is positive for all other values, it follows 
that V is a minimum at the position of equilibrium; the equilibrium is 
stable. 

Example 2. We shall now discuss a device known as Hooke's joint. 
This is used to transmit a torque or 
couple from one axis to another, 
inclined to the first. 

Figure 119 shows the essential 
features of the joint. AB is a shaft 
or axis, branching into the fork 
BCD] A'B' is another axis, with 
fork B'C'D'. These forks are con- 
nected by a rigid body composed of 
two bars CD, C'D', joined perpendic- 
ularly at their common center O. 
TJie lines AB, A'B' meet at when 
produced and are perpendicular to 
CD, C'D', respectively. There are 
smooth bearings at C, D, C', D', and 
the axes AB, A'B' are free to turn 
in smooth bearings at A and A'. 

Let I be a unit vector in the direc- 
tion AB and I' a unit vector in the 
direction B'A'. When a couple A/I is applied to ABCD, the system will 
move unless motion is prevented by other forces. We propose to calculate 
the couple M'V applied to A' B'C'D', which (together with MI and the 
reactions at the bearings) gives equilibrium. 

To solve such a problem, the general plan is as follows: (i) select general- 
ized coordinates qi, q*, - - for the system; (ii) calculate an expression of 
the form (10.708) for the work done in a displacement; (iii) equate the 
generalized forces to zero, and solve. 

We note that the fixed elements of the system are the lines AB, A'B' and 
the point O. Since ABCD can merely turn about AB, a single coordinate 
6 (the angle turned through) is sufficient to fix it. When ABCD is fixed, 
the line CD is fixed. Now the line C'D' must be perpendicular to both CD 
and A'B' (from the construction of the joint) ; hence C'D' is fixed, and so the 
angle suffices to fix, not only ABCD, but A' B'C'D' also. The system has 
one degree of freedom, and is a generalized coordinate. 

When 8 increases to 6 -f 50, A' B'C'D' turns through some small angle 50'; 
the displacements of ABCD, A'B'C'D', CDC'D' are as follows: 

ABCD-. a rotation 50 1, 

A'B'C'D': a rotation 50' I', 

CDC'D': a rotation dn = 5wii + 5nij + 5nk, 



FIG. 119. Hooke's joint. 



298 MECHANICS IN SPACE [Sue. 10.7 

where i, j, k are unit vectors along CD, C'D' and perpendicular to them. 
Since no work is done by the reactions at the bearings and the internal 
reactions at C, C", D, />', we have [cf. (10.704)] 

(10.715) 6W m Ml dO 1 + M T 50' I' 

= M SB + M ' 60'. 

We must now find 69' in terms of 60. The point D belongs to two rigid 
bodies ABCD and CDC'D'. Equating the two expressions for its dis- 
placement fcf (10.501)), we have 

(10.716) 60 1 X ai - 6n X ai, 

where a = OD OD'. Similarly, by considering the displacement of D' t 
we find 

(10.717) 60' I' X aj - 5n X aj. 
Now, resolving along i, j, k, we have 

I = sin < j + cos </> k, I' sin tf>' i + cos #' k, 

where <f>, <J>' are the angles between k and AB, B'A', respectively. Sub- 
stituting these values for I and I' in (10.716) and (10.717), we obtain 

l cos * 80 = dn *> ~ sin * 5 * " 5n 

| cog ^ ^, _ fi ^ _ gm ^ 5(?/ _ gn ^ 

Equating the expressions for 5^a, we obtain 

(10.719) 60 f - cos sec </ ^. 
Substituting this value in (10.715), we get 

(10.720) 6W - (Af -f M' cos sec </>') 5^, 
and so the single generalized force is 

6 M -f Jf ' cos sec <'. 

For equilibrium, this must vanish, and so the couple required to hold 
A'B'C'D' is MT, where 

M' - - 3f sec cos '. 

The fraction of the torque M transmitted through the joint is sec <t> cos 0'. 

Example 3. In Fig. 120, A B represents a shaft free to turn about a hori- 
zontal axis L, perpendicular to AB at A ; DE represents a heavy bar threaded 
on the shaft AB and perpendicular to it. The pitch of the thread is p, so 
that, when DE turns through an angle about AB, E moves along AB 
through a distance p<f>. The points C, C' are the centers of gravity of AB, 
DE, respectively. We wish to find the possible positions of equilibrium of 
this system under gravity, all friction being neglected. 

We start with a standard configuration in which AB is vertical and DE 
lies in the vertical plane II perpendicular to L at A. Let XQ denote the 
distance AE in this position. We pass to a general configuration by nwing- 



SBC. 10.7] 



STATICS IN SPACE 



299 



ing AB through an angle and turning DE to make an angle 4> with II. The 
angles and <t> are generalized coordinates. In terms of them, the potential 
energy is 

(10.721) V - -wa cos 9 - W[(XQ 4- p<) cos + 6 sin 9 cos <], 

where a AC, 6 = EC' and 10, W denote the weights of AB, DE, respec- 
tively. 

The conditions of equilibrium are 



(10.722) 



wasm 6 + W(XQ -f P^) sin Wb cos cos < = 0, 



dV 

_ =, _ PFp cos + Wb sin sin <j> 

i O<f> 



0. 



With numerical values for the constants, this equation can be solved graph- 
ically or otherwise. In terms of <, 6 is given by 



Elimination of gives for < the equation 
(10.723) Wb 2 sin 20 - 2wpa -f 



(10.724) 



tan0 



Some interesting results can be deduced without solving (10.723). If p is 
small, a position of equilibrium occurs for some small value of <, i.e., with 




Fio. 120.- 



-A bar ED is threaded on a shaft AB, which can turn about a horizon- 
tal axis L. 



DE close to II. For this position, tan 6 is finite, since p and < arc small of 
the same order. Another position occurs near = $TT; for this, tan is 
small, and so AB is nearly vertical. If p is large, the right-hand side of 
(10.723) may exceed Wb* for all 4>; in this case, there is no position of equilib- 
rium, and the bar DE simply runs down the shaft AB, turning as it goes. 

Example 4. The last two examples considered above are three-dimen- 
sional in character. We conclude with an application of the methods of 
work and energy to a two-dimensional system with many degrees of freedom. 



300 



MECHANICS IN SPACE 



[SEC. 10.7 



Consider a chain of n equal uniform rods, smoothly jointed together and 
suspended from one end A\. (Figure 121 shows the case n = 5). A hori- 
zontal force P is applied to the other end A n +i of this chain. It is required 
to find the equilibrium configuration. 




FIG. 121. A chain of rods pulled by a horizontal force. 

As generalized coordinates, we take the inclinations (to tho downward 
vertical) 0i, 2 , O n of the several rods in order. If each rod has length 2a 
and weight w, the potential energy in a general configuration is 

V = wa cos 0i w(2a cos 0\ + a cos 2 ) 
w(2a cos 0i H- 2a cos 2 + 
= - wa[(2n - 1) cos 0i -f (2n - 3) cos 2 



2a cos n . j -fa cos 0) 
-f 3 cos 0_j + cos 0]. 



In a small virtual displacement the work done by gravity is 

-87 = -t0a[(2n - 1) sin 0i 50i + (2n - 3) sin 2 80 2 + sin 50 n ], 

The work done by the force P is the product of P and tho horizontal dis- 
placement of *4nfi, i.e., 

P5(2a sin 0i + 2a sin 2 + + 2a sin n ). 
Adding these two expression, wo find, for tho tot.il work done, 
dW = [2P cos 0i (2n - \}w sin t ]a 50i 



(10.725) 



+ [2P cos 2 - (2n - 3)iv sin 2 ]a 50 2 

+ - - 

-f I2/ 3 cos tt ^ sin n ]a 50 B . 



This must vanish for arbitrary values of 50i, 60 2 , 50 n , if tho displace- 
ment is from a position of equilibrium. Hence, equating to zero the brackets 
on the right of (10.725), we find 



SEC. 10.8] STATICS IN SPACE 301 

2P 
/ tan 0i = - 

2P 
(10.726) 



* fl 2/> 

tan U = 

w 

These equations give the inclinations of the rods to the downward vertical 
in the equilibrium configuration, the tangents form a harmonic progression. 

10.8. SUMMARY OF STATICS IN SPACE 

I. Conditions of equilibrium. 

(a) For a single particle (necessary and sufficient) : 

(10.801) P = 0, 

or 

* 

(10.802) X = 0, Y = 0, Z = 0. 

(b) For any system (necessary), or for a rigid body (necessary 
and sufficient) : 

(10.803) F = 0, G = 0. 

(F = total force, G = total moment.) 

(c) For any system with workless constraints (necessary and 
sufficient) : 

(10.804) dW = 0, (5W = work done by applied forces). 

II. Equipollence. 

(a) Conditions of equipollence : 

(10.805) F = F, G = G'. 

(6) Any system of forces can be reduced to a force F at an 
assigned point, together with a couple G. If G = pF, the 
reduced system is a wrench. 

III. Displacements of a rigid body. 

(a) Finite displacements: 

(i) Any displacement of a rigid body with a fixed point is 
equivalent to a rotation n (Euler's theorem). 



302 MECHANICS IN SPACE [Ex. X 

(ii) A general displacement is equivalent to a translation s, 

followed by rotation n. 
(6) Infinitesimal displacements: 
(i) Infinitesimal rotations compound vectorially. The order 

of application is immaterial 

(ii) For a rigid body with a fixed point, the displacement 
of a particle of the body is 

(10.806) 5n X r. 

(iii) In general, the displacement of a particle of the body is 

(10.807) Ss + 6n X r. 

IV. Work and potential energy. 

(a) Work done on a particle: 

(10.808) BW = P - 5s = X 8x + Y By + Z Sz. 

(b) Work done on a rigid body: 

(10.809) 8W = F 5s + G 5n. 

(c) Work done on a general system: 

(10.810) 8W = Qi fyi + Q 2 fy 2 + - + Q n 8q n . 

(d) Work done on a conservative system: 

n *y 

(10.811) dW = -57 = - J) Sp Sqr. 



EXERCISES X 

1. A force with components (-7, 4, -5) acts at the point (2, 4, 3). 
Find its moment about the origin. Find also its moment about the line 

x = y - z, 

the positive sense on the line being that in which x increases. 

2. A rigid body is acted on by a force with components (1, 2, 3) at a 
point (3, 2, 1) and by a force with components ( 1, 2, 3) at a point 
(-3, 2, 1). Give the components of the equipollent force and couple 
at the origin. 

3. A particle of weight w is placed on a rough plane inclined to the hori- 
zontal at an angle a. If the coefficient of friction is 2 tan a, find the least 
horizontal force across the plane which will cause the particle to move. 
Determine the direction in which the particle moves. 



Ex. X] STATICS IN SPACE 303 

4. A tripod consisting of three uniform rigid legs, each of length 2a 
and weight w, supports a camera of weight W, the legs being smoothly 
jointed together at the top. The tripod stands on a rough horizontal plane 
(coefficient of friction /*), the feet forming an equilateral triangle. Find 
an expression for the greatest length of a side of this triangle consistent with 
equilibrium. 

5. Prove, by the principle of virtual work, that for an inextensiblc 
cable (either free or in contact with a smooth surface) T V constant, 
where T is the tension and Yds the potential of the external force acting on 
an element ds of the cable. 

6. A force with components (3, 5, 6) acts at a point with coordinates 
(1, 2, 3), and a force with components ( 8, 2, Z) acts at a point with 
coordinates (4, 6, 7). If the pair of forces has no resultant moment about 
the x-axis, find Z. 

7. Determine the pitch of the wrench equipollent to two forces of magni- 
tudes P, Q, inclined to one another at an angle a, the shortest distance 
between their lines of action being c. 

8. A square gate ABCD, of weight W and edge a, has hinges at B and C, 
the line BC being vertical with B on top. The hinge at C can support a 
downward thrust, but that at B merely supplies a vertical axis of rotation. 
The wind blows on the gate, exerting a uniform pressure p. The gate is 
kept in position by a light rope attached to the outer upper corner A and to a 
point E on the ground, where CE a and CE is a perpendicular to the gate. 
Find in terms of W, p, a the tension in the rope and the magnitude of the 
reaction at each of the hinges. 

9. Three identical spheres lie in contact with one another on a horizontal 
plane. A fourth identical sphere rests on them, touching all three. Show 
that the coefficient of friction between the spheres is at least (-\/3 \/2) 
and that the coefficient of friction between each sphere and the plane is at 
least (V3 - V2)/4. 

10. Denoting by 0, $ the usual polar angles, find the polar angles of the 
axis of an infinitesimal rotation equivalent to three infinitesimal rotations, 
all of the same magnitude, with axes whose polar angles are 

(0 - 60, 4> - 45), (0 = 120, $ = 135), (0 = 60, = 225). 

11. A rigid body receives in succession three rotations about three 
mutually perpendicular intersecting lines fixed in space, each rotation being 
through a right angle and the senses being cyclic. Find the axis and magni- 
tude of the single equivalent rotation. 

12. A rigid body receives a finite translation and a finite rotation (through 
an angle B) about an axis D perpendicular to the translation. Show that the 
resultant displacement is equivalent to a rotation (through an angle 0) 
about an axis parallel to D, the position of this axis depending on the order 
in which the translation and rotation are applied. 

13. Show that any finite displacement of a rigid body is equivalent to a 
screw, i.e., a translation and a rotation about an axis parallel to the transla- 
tion (Chasles' theorem). 



304 MECHANICS IN SPACE [Ex. X 

14. In coining to rest on a slippery road, the wheel of a car travels 10 foot 
forward and 2 feet sideways, at the same time turning through an angle of 
180 about its axle. Locate the axis of the equivalent screw displacement. 

15. Show that, in general, a force system may be reduced to a force acting 
along any given line together with another force. (The lines of action of the 
two forces are said to be conjugate ) 

16. A rigid body is acted on by a force F at and a couple G. P is an 
assigned point, with position vector r relative to 0. Show that there is 
a single infinity of lines through P about which the force system has no 
moment; show that these lines lie in a plane and find, in Cartesian coordi- 
nates, the equation of this plane. (The lines are called null hnes, and 
the plane a null plane.} 

17. Show that, if a rigid body is in equilibrium under the action of four 
forces, the invariant (F G) of any two is equal to the invariant (F G) of 
the other two. Show also that the invariant (F G) of any three of the 
forces is zero. 

18. A heavy uniform inextensible cable hangs in contact with a smooth 
right-circular cone of semiverticul angle a, the axis of the cone being vertical. 
Prove that the cable hangs in a curve satisfying the equation 

(ID* + z * sin2 a = z * (A + Bz}2 > 

where z is the depth below the vertex of the cone, <f> is the azimuthal angle, 
and A, B are constants. 

19. In a Hooke's joint (Fig 119) force systems (F, G) and (F', G') (includ- 
ing the reaction of the bearings at A, A') act on the parts A BCD, A'B'C'D', 
respectively, being taken for base point. For equilibrium, show that 
couples G, G' must both be perpendicular to the plane CDC'D*. 

20. There are two identical rough stones, each being an oblate spheroid 
of serniaxes a, b (a > b). One is laid on a horizontal iloor and the other 
balanced on top of it, the axes of symmetry being vertical. Confining atten- 
tion to displacements in a vertical plane through the axis of symmetry, show 
that the equilibrium is stable if a 2 > 36 2 . 

21. Discuss the stability of any four successive positions of equilibrium 
for the system shown in Fig. 120. Consider only the case where p is small 
in comparison with a and 6. 

22. A force 1 system is equipollent to a force F at and a couple G; another 
force system is equipollent to a force F' at O and a couple G'. Prove that, 
if the axes of the equivalent wrenches intersect, then 

F G' + F' G - (p + />')F F', 

where p, p' are the pitches of the wrenches. 

23. In a Hooke's joint (Fig. 119) let the angle 0, through which ABCD 
is turned about AB, be measured from zero when CD lies in the plane of 
AB and A'B'. Denoting by a the angle between AB and B'A', find the 
angles <f> and ' in terms of and a. Check your answers by considering 
the special cases: (i) 0, (ii) 6 %*. 



CHAPTER XI 

KINEMATICS. KINETIC ENERGY ANH ANGULAR 
MOMENTUM 

We now approach the study of dynamics in space. We shall 
require 

(i) a simple way of describing the motions of particles and of 
rigid bodies; 

(ii) methods of calculating kinetic energy and angular 
momentum. 

These items belong to kinematics (if we understand the word to 
include mass as well as motion) and form the subject matter of 
the present chapter. 

11.1. KINEMATICS OF A PARTICLE 

Let Oxyz be rectangular axes fixed in a frame of reference and 
I, J, K unit vectors along them. For any particle, with coordi- 
nates x, y, z, we define the following vectors (cf. Sec. 1.3): 



(11.101) 



Position vector: r = xl + yj + zK, 
Velocity: q = -r = .rl + yj + zK, 

Acceleration: f = ~ = jcl + yj + zK. 
at 



The simplest way of describing the motion of a particle is to 
give the vector function r(J). For, when r is known as a vector 
function of the time t (i.e., when x, y, z are known as scalar 
functions of t), we can trace the path of the particle and 
find its velocity and acceleration at any instant by differentiation. 

We frequently require expressions for the components of 
velocity and acceleration in directions other than I, J, K. Two 
particular resolutions of these vectors will now be considered. 

Tangential and normal components of velocity and acceleration. 

Figure 122 shows the path C of a moving particle A\ AQ is 
a fixed point on C. The arc length AA Q is denoted by s. From 

305 



306 



MECHANICS IN SPACE 



[SBC. 11.1 



(11.101), we see that the vector di/ds has components dx/ds, 
dy/ds, dz/ds along I, J, K; it is the unit tangent vector to C 
at A and will be denoted by i. 

For the velocity of A we have 



K 




<" t-- 



si. 



Hence, the velocity of a .particle is 
directed along the tangent to its path, 
and has magnitude s. 

For the acceleration we have 



x dq .. , . 

f = Tt= sl + s 

But, by (10.211), 



di 



FIG. 122. A particle moving in 
space. 



-_-==, 

as p 



where j is the unit principal normal vector and p the radius of 
curvature of C at A. Hence, 

* 2 

(11.103) f = 6 : i + -j, 

and so we may state: The acceleration of a particle lies in the 
osculating plane to its path; the components in the directions of 
the tangent and principal normal are s and s 2 /p, respectively. These 
results should be compared with those for the corresponding two- 
dimensional case (cf. Sec. 4.1). 

Components of velocity and acceleration in cylindrical coordi- 
nates. 

In Fig. 123, A is the position of a particle at time t and M 
the foot of the perpendicular from A on the plane Oxy. The 
polar coordinates (R, <j>) of M, together with the ^-coordinate 
of A, are the cylindrical coordinates (R, <j>, z) of A. Let i, j, k 
be unit vectors at A in the directions of the parametric lines of 
these coordinates (i.e., those directions in each of which just 
one of the three coordinates R, <, z increases, the other two 
remaining constant). We wish to find the components of the 
velocity and acceleration of A along i, j, k. 



SEC. 11.1] 



KINEMATICS 



307 



The vector k is constant in magnitude and direction. The 
directions of i and j do not depend on R and z; they are, however, 
dependent on <t>. As in Sec. 4.1 (where r, correspond to 
R, 4>), we have 



(11.104) 
Since 



d<t> 




i 
FIG. 123. Cylindrical coordinates. 

we obtain, on differentiating r with respect to t and using (11.104), 

(11.105) q = ~ = Ri + Rfo + *k, 
A second differentiation with respect to t gives 

(11.106) f = (R - RW\ + jjj t (# 2 <ttJ + *k- 

From these equations, we can read off the components of the 
velocity (q) and the acceleration (f) in the directions of i, j, k, 
when required. 

Composition of velocities and accelerations. 

We often need to connect the velocities (or accelerations) 
of a particle relative to two different frames of reference, S and 
S'. We shall here think only of the case where there is no 
relative rotation of the frames. Let be a point fixed in S and 



308 MECHANICS JN SPACE [Sue. 11.2 


O r a point fixed in 8'. A particle A has position vectors r = OA 

and r' = 0' A ; they are connected by 

(11.107) r = r + r', 
> 

where r = 00'. Differentiation gives 

(11.108) q = qo + q', f = f o + f , 

where q, f = velocity and acceleration of A relative to S, 
q', f = velocity and acceleration of A relative to S', 
q , fo = velocity and acceleration of S' relative to S. 

The equations (11.108) give the laws of composition of velocities 

and accelerations. 

11.2. KINEMATICS OF A RIGID BODY 
Motion of a rigid body with a fixed point. 

Consider a rigid body constrained to rotate about a fixed 
point O. Let t\, fa be two instants; in the time interval fa t\ 
the body receives a displacement which is equivalent (cf. Sec. 
10.5) to a rotation n about 0. If we keep ti fixed and let fa 
approach ti, the direction of n will approach some limiting 
direction, which we denote by the unit vector i. The ratio of 
the angle of rotation n to the time interval fa fa will approach 
a limiting value co. The vector G> = coi is called the angular 
velocity of the body at the instant t\. At this instant the body is 
rotating about a line through in the direction of <o; this line 
is called the instantaneous axis of rotation. The rate of turning 
is co radians per unit time and is a rotation in the positive sense 
about the instantaneous axis. 

In an infinitesimal time dt the body receives an infinitesimal 
rotation <o dt; and so the displacement of a particle of the body 
is, by (10.501), 

dr = (o dt X r, 

where r is the position vector relative to 0. The velocity of 
this particle is 

(11.201) ' q = J = X r. 

This formula gives the velocity of any particle of the body in 
terms of the angular velocity vector <o. Thus, if o> is known as 



SBC. 11.2] KINEMATICS 309 

a vector function of the time, we can find the velocity of any 
particle at any time; in other words, the single vector function 
o() suffices to describe the motion. 

As the body turns about 0, the instantaneous axis (determined 
by <o) will occupy different positions in the body. Since this 
axis always passes through 0, its locus in the body is a cone with 
vertex 0; it is called the body cone (or polhode cone). Similarly, 
the locus of the instantaneous axis in space is another cone with 
vertex 0; it is called the space cone (or herpolhode cone). 

A rigid body moving parallel to a fundamental plane may be 
regarded as a body turning about a point at infinity. In this 
case the body and space cones become cylinders; their inter- 
sections with the fundamental plane are the body and space 
centrodes of our earlier theory (cf. Sec. 4.2). 

We saw in Sec. 4.2 that, in the? motion of a rigid body parallel 
to a plane, the body centrodo rolls on the space centrode. Simi- 
larly, in the motion of a rigid body with a fixed point, the body 
cone rolls on the space cone. To establish this result we must 
show that: 

(i) the body cone touches the space cone; 

(ii) tho particles of the body on the line of contact of the cones 
are instantaneously at rest. 

Let OA be the position of the instantaneous axis of rotation at 
some instant. It is a generator of the fixed space cone and also 
of the moving body cone. After an infinitesimal time dt, 
another generator OB of the body cone comes into coincidence 
with a generator OB' of the space cone. But the displacement 
in time dt is an infinitesimal rotation of magnitude co dt about OA, 
and so the angle between the planes OA B, GAB' is an infinitesimal 
angle. Since these planes represent the tangent planes to the two 
cones along the generator OA, it follows that the tangent planes 
cannot cut at a finite angle; the cones must therefore touch. 
Since all particles of the body on the instantaneous axis OA are 
instantaneously at rest, the second of the above conditions is also 
satisfied, and the result is established. 

The components of angular velocity in terms of the Eulerian 

angles. 

In Sec. 10.6 we defined the Eulerian angles 0, <, ^; they 
describe (relative to a fixed triad I, J, K) the position of a triad 



310 MECHANICS IN SPACE [Sue. 11.2 

of unit orthogonal vectors i, j, k, fixed in a rigid body turning 
about a point (Fig. 118). The motion of the body is deter- 
mined when 6, <t>, $ are known as functions of the time t; but 
this motion can also be described by the angular velocity o>(). 
We write 



and seek expressions for i, 2 , w 3 in terms of 0, <, t, and their 
rates of change. 

In an infinitesimal time dt the body receives the rotation 
o dt. But, by (10.607), this rotation is 

(sin ^ d6 sin 6 cos ^ d<t>)\ -t- (cos ^ d0 + sin sin ^ d<t>)j 

+ (cos d< + 



where d0, d<, d\fr are the infinitesimal increments in 0, <, ^ in 
time dt. Equating this expression to <*di and dividing by dt, 
we have 

!coi = sin ^ 6 sin 6 cos ^ <, 
<o 2 = cos ^ + sin 6 sin ^ <, 
w 3 = cos <^ + \l/. 

These equations give the components of angular velocity when 
the motion is known, i.e., when 9, <, \f/ are known as functions of 
the time. Conversely, when the components of <> are known at 
any time t, we can solve the above equations for 0, <, ^ as func- 
tions of t and so determine the motion. 

Exercise. Find the components of <*> on the fixed triad I, J, K (Fig. 118) 
in terms of 9, <t>, ty and their rates of change. 

General motion of a rigid body. 

Let us consider a rigid body moving in a general manner. We 
select a particle A of the body as a base point and denote its 
velocity by q^. In an infinitesimal time dt, the displacement 
of the body is equivalent to a translation q^ dt, and a rotation 
dn about A (cf. Sec. 10.5). By (10.502), the displacement of 
any particle B of the body is 

q^ dt + dn X r, 

where r = AB. Hence, for the velocity of B we have 
(11.203) q = q^ + *> X r, 



SBC. 



KINEMATICS 



311 



where co = dn/dt. We observe that this velocity consists of two 
parts: (i) the velocity q^ of the base point, and (ii) the velocity 
of B relative to A, viz., <o X r. It is clear that the velocity of B 
relative to A is precisely the same as if the body were turning 
about A (as a fixed point) with angular velocity <o. 

If we alter the base point A, the translation q A dt is changed, 
but the rotation da remains the same. It follows that the 
vector o> pertains to the motion of the body as a whole; it is the 
angular velocity of the body and is to be regarded as a free vector, 
since it does not depend on our choice of base point. The 
equation (11.203) gives the velocity of any point of the body 
when the angular velocity <o and the velocity q^ are known; 
thus, the two vectors o> and q A completely describe the motion. 

When CD and q A are known as vector functions of the time, 
we have a picture of the motion at any instant. From the 
velocity q of any particle J5, as given 
by (11.203), its acceleration f can be 
found by differentiation; thus, 




Fia. 124. A wheol rolling on a 
straight track. 



We have 



Here d^A/dt is the acceleration f A of 

the base point A ; it depends solely 

on the motion of A and not on the 

angular velocity. As for the last 

term, di/dt is the velocity of B 

relative to A] therefore, by (11.201), it equals c* X r. 

then 

(11.204) f = i A + ~ X r + <o X (o> X r). 



Example 1. As a simple illustration, let us consider a circular wheel 
rolling with constant speed along a straight level track (Fig. 124). We take 
as base point the center C of the wheel and denote its velocity by V. This 
vector is constant; it lies in the plane of the wheel and is horizontal. The 
angular velocity <> of the wheel is a vector perpendicular to its plane; it also 
is a constant vector. By ( 1 1 .203) , a particle B of the wheel has velocity 

V + o X r, 
> 

where r = CB. Since <* X r is a vector perpendicular to *>, this velocity 
lies in the plane of the wheel a fact which is intuitively obvious. Since 



312 



MECHANICS IN SPACE 



[SBC. 11.2 



and V arc constant vectors, the acceleration of B is, by (11.204), 



f = <a X (<* X r) = 



r) - r 2 = -r 2 . 



Thus, each particle of the wheel has an acceleration of magnitude ro> 2 , 
directed toward C. 

Example 2. As a second illustration, let us consider the motion of the 
propeller of an airplane making a turn. In particular, let us see how the 
velocity and acceleration of the tip of the propeller may be found. 

For simplicity, we shall suppose that the center of the propeller describes 
a horizontal circle C with constant speed F; let & be the radius and A the 
center of C. Figure 125 shows the position of the propeller when the line 
from the center to the tip B makes an angle with the vertical. 




Flo. 125. Motion of an airplane propeller. 

Let i, j, k be a triad of unit orthogonal vectors at O; i points along AO, 
k points vertically upward, and j completes the triad. The vector j is 
clearly the unit tangent vector to C at 0. As a base point for the descrip- 
tion of the motion, we take the point 0; its velocity is Vj. The angular 
velocity o of the propeller consists of two parts : 

(i) an angular velocity or spin sj (where s = 6), imparted by the engine; 

(ii) an angular velocity (V/6)k, due to the turning of the airplane. 
Hence, 



The second part of G> arises from the fact that, in time 2irb/Vj the airplane 
(and the axis of the propeller) would turn through an angle 2?r about the 
vertical. 

From (11.203), we have, for the velocity of any point of the propeller 
(position vector r relative to 0), 

q = Vj + u X r. 



SBC. 11.3] KINEMATICS 313 

In particular, the velocity of B is 

(11.205) q fl = Vj + (sj + ~ k\ X (a sin i + a cos k) 
= v cos i + (l + j sin Wj - t; sin k, 

where a = OB and t> = sa, the speed of the tip relative to the airplane. 
Hence the absolute speed is given by 

9fl = 2 + V 2 (l + | sin 0) 2 . 

Actually, a/6 will be small, and so g| = w 2 -+- F 2 , approximately. If the 
trigonometrical term is retained, qn takes maximum and minimum values 
when the propeller is horizontal. 

The acceleration of B may be found by differentiating (11.205). We shall 
assume that s is constant; then the scalars , V, b and the vector k are con- 
stant, whereas the scalar and the vectors i and j are variable. To find 
di/dt and dj/dt, we note that i and j may be regarded as the position vectors 
of points fixed in a body, turning about O with angular velocity (F/&)k. 
Hence, by (11.201), 



It is left for the reader to verify that the acceleration of B is 

(v 2 V 2 V 2 a \ 2vV v 2 

sin H --- h sin B J i + -r- cos j -- cos k. 
a b b 2 / b a 

Under normal circumstances, the terms in v 2 /a far exceed the other terms in 
magnitude, and so the acceleration is due almost entirely to the spin of the 
propeller. 

11.3. MOMENTS AND PRODUCTS OF INERTIA 

The moment of inertia of a system was defined in Sec. 7.1. 
For a particle of mass m distant p from a line L, the moment of 
inertia about L is mp 2 . For a system of particles, the moment 
of inertia is the sum of the moments of inertia of the several 
particles. 

We now define products of inertia. Let P, Q be two planes, 
and let p, q denote the perpendicular distances from them of a 
particle of mass m. The distance is counted positive or negative 
according as the particle lies on one side or the other of the 
corresponding plane. The product mpq is called the product 
of inertia of the particle with respect to the planes P, Q. For a 
system of particles, the product of inertia is the sum of the 
products of inertia of the several particles. 



314 



MECHANICS IN SPACE 



flc. 11.3 



Let Oxyz be rectangular axes; the moments of inertia of a 
system of particles about the axes Ox, Oy, Oz are, respectively, 



(11.301) A = 2m(?/ 2 + z 2 ), B = Sm(z 2 



+ z 2 ) 
C = 



Here m is the mass of a typical particle, x, y, z are its coordi- 
nates, and the summation extends over all particles of the 
system. The products of inertia with respect to the coordinate 
planes, taken in pairs, are 

(11.302) F = Swyz, G = 2mzx, H = Zmxy. 

For a continuous distribution of matter the summations are 
replaced by integrations, the mass m being replaced by the mass 
pdr (p = density) of a small volume element dr. 

It is a remarkable fact that, 
when A, B, C, F, (7, H are known, 
we can find the moment of inertia 
/ of the system about any line 
through 0. To see this, we 
recall that, by definition, 




FIG. 126. The moment of inertia 

about the line L is required. 



of the vector product 



where p is the perpendicular 
distance of a typical particle P 
(mass m) from the line L (Fig. 
126). Now p OP sin 0, where 
Q is the angle between OP and 

T ^i i xi -j. i 

L] thus, p equals the magnitude 
X r, where 3* is a unit vector along L 



and r = OP. The components of ^. are the direction cosines 
a* )3, 7 of L, and the components of r are the coordinates x, y, z 
of P. Hence, the components of ^ X r are 

$z yy, yx az, ay fix. 

Thus, since p is the magnitude of the vector with these com- 
ponents, we have 

(11.303) 7 = 2m[(0* - T2/) 2 + (7* - **) 2 + (0 - 0*) 2 1 * 



SEC. 11.3] KINEMATICS 315 

This gives 7 in terms of A, B, C, F, , H, and the direction cosines 
ofL. 

When A, B, C, F, 6r, // are known for any set of rectangular 
axes through the mass center, we can find the moment of inertia / 
of the system about any line L very easily. This is done in 
two steps: 

(i) use (11.303) to find the moment of inertia 7 about a line 
through the mass center parallel to L; 

(ii) apply the theorem of parallel axes (cf. Sec. 7.1) to find /. 

If A, B, C, F, G, H are known for a point other than the mass 
center, we can find I in a similar manner, but two applications 
of the theorem of parallel axes arc required. 

The momental ellipsoid. 

By varying a, 0, 7 in (11.303), we obtain the moments of 
inertia about all lines through 0. Let us measure off, along 
each line through 0, a distance OQ = l/-\/I, where / denotes 
the moment of inertia about the line in question. The locus of 
Q has the equation 



(11.304) Ax 2 + By* + Cz* - 2Fyz - 2Gzx - 2Hxy = 1. 

This is the equation of a quadric surface with center 0; in 
general, it is a closed surface, since / does not vanish for any 
line.* Hence, (11.304) is the equation of an ellipsoid; it is called 
the momental ellipsoid at 0. 

When the equation of the momental ellipsoid at a point is 
known, we find the moments and products of inertia with 
respect to the axes of coordinates by inspecting the coefficients 
in this equation. Under a rotation of axes from Oxyz to Ox'y'z', 
the equation of the momental ellipsoid changes from (11.304) to 

AV 2 + B'y' z + C'z' 2 - ZF'y'z' - 2G'z'x' - ZH'x'y' = 1. 

The coefficients A 1 , B', C', F', G', //' give the moments and 
products of inertia for the new axes. 

The quadric represented by the equation (11.304) sums up 
the inertial properties of the system with respect to axes through 
the origin. The form of the equation changes (in the sense 
that the values of the coefficients change) when we rotate the 

* There is only one exceptional case. If all particles of the system lie on 
a line L, then / - for L; the quadric is then a circular cylinder with axis L. 



316 MECHANICS IN SPACE [SEC. 11.3 

coordinate axes, but the quadric itself remains an invariant 
model of the inertial properties. The representation of physical 
properties by means of a quadric surface is of frequent occur- 
rence it is used in elasticity, hydrodynamics, and other branches 
of applied mathematics. Since the coefficients in the equation 
of the quadric change when we change the axes, they cannot bo 
called scalars, in the sense that mass is a scalar. Nor are they 
components of a vector. The whole set of six coefficients, or 
more precisely the array 

A -H -G 

(11.305) -// B -F 

-0 -F C 

is called a tensor. This is the simplest example of the con- 
cept which has played such an important part in the theory of 
relativity. 

An array is also called a matrix. Just as we use a single letter 
to denote a vector (which may be regarded as a matrix with 
three elements), so we may denote a matrix by a single letter. 
The operations of algebra may be applied to matrices, yielding 
a compact and powerful notation in mechanics. * 

Existence of principal axes and moments of inertia. 

Equation (11.303) gives us the moment of inertia about any 
line L through in terms of the direction cosines of L and the six 
coefficients shown in the array (1 1.305). We shall now show 
that the number of coefficients may be reduced from six to three 
by making a suitable choice of the axes Oxyz. 

Let Oxyz be any axes. Then 7, as given by (11.303), attains 
its maximum value for some line LI. Let this maximum be I\. 
Let us take axes Ox'y'z' such that Ox r coincides with LI; we leave 
the directions of the other two axes unspecified for the present, 
except for the conditions that they shall be perpendicular to Ox' 
and to one another. Then, for any line L, the moment of 
inertia is 

I = A V 2 + B'p"> + CV 2 - 2F'p'y' - 2G'y'a' - 277V/3', 

* See R. A. Frazer, W. .1. Duncan, and A. R. Collar, Elementary Matrices 
(Cambridge University Press, London, 1938). 



SEC. 11.3] KINEMATICS 317 

where A', B' ', C", F', G', 7T, are the moments and products of 
inertia for the axes Ox'y'z' ', and a', /3', 7' are the direction cosines 
of L relative to Ox'y'z'. If we put a' = 1, 0' = 7' = 0, then L 
coincides with LI, and so A' = 7i, the maximum moment of 
inertia. 

We shall now show that the vanishing of G' and H r is a neces- 
sary consequence of the fact that / is a maximum for of = 1, 
p = y = 0. Since a' 2 + 0' 2 + 7 /2 = 1, we can write 

7 - 7 X = -2a'(G'y' + H'ff) + (B' - 7i)/3' 2 

+ (G" - 7ih' 2 ~ 2F'0Y. 

If we take a line L near LI, a/ will be nearly unity and /3', 7' will 
be small. If at least one of G' , H f is different from zero, we can 
choose /3', 7' (reversing one or both signs if necessary) so that 
(G'y' + H'0') is negative. But, since /3' and 7' arc small, the 
sign of the right-hand side of the above equation is determined 
by the first term. Therefore, 7 7i may be made positive. 
But this is impossible since I\ is the maximum of 7. Therefore 
the hypothesis we made about G' and H' is false, and we conclude 
that G f II' = 0, so that for any line L, 

7 = 7i' 2 + B'p* + CV 2 - 2* V 0Y- 

This is true for all axes Ox'y'z' such that Ox' coincides with LI. 
Let us now subject Oy r to the condition that, of all lines per- 
pendicular to Ox' , Oy' has the maximum moment of inertia, say 
7 2 . Then we have B' 1^ and so for any line L, 



For lines L perpendicular to Ox' we have a' = 0, 0' 2 + 7 /2 = 1, 
and so 

/ - / 2 = -2F'/3Y + (C' - 7 2 ) T /2 . 



If we take a line near Oy 1 ', ft' will be nearly unity and 7' will be 
small. It is evident that, if F' docs not vanish, we can make 7 
greater than 7 2 , which is impossible since 7 2 is a maximum. 

Therefore F f 0, and we have for any line L, 
7 = 7!' 2 + 7 2 /3' 2 + CY 2 . 
Substituting 7 /2 = 1 - a' 2 - /2 , we get 

/ - C" = (/i - <7>' 2 + (7 2 - C")0 /2 ^ 0, 



318 MECHANICS IN SPACE [SEC. 11.3 

and so the third moment of inertia C" is the least of all moments 
of inertia for lines through 0. 

We may sum up as follows, simplifying the notation: It is 
always possible to choose rectangular axes Oxyz such that the moment 
of inertia I of a system about a line L through is given by 

(11.306) I = Aa* + Bp* + Cy*, 

where a, /?, 7 are the direction cosines of L relative to Oxyz. These 
axes are called principal axes of inertia at 0, and the moments of 
inertia A, B, C about them are called principal moments of 
inertia. The planes defined by the principal axes are called 
principal planes; for any pair of principal planes, the product of 
inertia vanishes since F, G, and H are absent from (11 .306). For 
principal axes, the equation of the momental ellipsoid is 

(11.307) Ax 2 + By* + Cz* = 1. 

Exercise. Find principal axes of inertia for a thin straight uniform rod 
at its middle point. 

General method of finding principal axes and moments of inertia. 

The principal axes and moments of inertia have been shown to 
exist. We shall now show how to find them, starting from 
general axes Oxyz with moments and products of inertia A, B, 
C, F, G, H. Let Ox'y'z' be the principal axes and A 1 , B', C' the 
principal moments of inertia. Any point P has two sets of 
coordinates, (x, y, z) and (a;', y' ', z'), according to the axes which 
are used. One set of coordinates are linear functions of the 
other, such that 

x* + y* + z 2 = x'* + y'* + z' 2 

for every point P. Also, for every point P, we have 
Ax* + By* + Cz* - 2Fyz - 2Gzx - 2Hxy = A'x'* 

+ By* + c f z'*, 

since each side represents the moment of inertia about OP, 
multiplied by OP 2 . Therefore, no matter how the constant K 
is chosen, we have the identity 

Ax* + By* + (Jz* - 2Fyz - 2Gzx - 2Hxy - K(x* + y* + z*) 
= AV 2 + By* + C'z r * - K(x f * + y'* + z'*). 

Let us denote each side of this identity by $. Consider the 



SEC. 11.3] KINEMATICS 319 

equations 



Explicitly, these equations read 

(A' - K)x' == 0, (' - K)y' = 0, (C" - K)z' = 0. 

Rejecting the trivial solution x' = ?/' = z' = 0, we must choose 
/f equal to A' r , Z? 7 , or C". We have then the following three 
solutions : 

K = A', a/ arbitrary, y' = 0, z r = 0; 
7f = ', x' - 0, y 1 arbitrary, z' = 0; 
K = C", a:' = 0, y' = 0, 2:' arbitrary. 

Thus (11.308) have nontrivial solutions provided K is equal to 
one of the principal moments of inertia; the corresponding values 
of x' y y' y z' give the principal axes. 
Now 



_ 

dx dx f dx dy' dx dz'dx' 

and similar equations could be written for d$/dy and d$/dz. 
Therefore, (11.308) imply that 



If K", x', y', z' are chosen as above, (11.308) are satisfied, and 
therefore (11.309) are satisfied. Explicitly, (11.309) read 

( (4 - K)x - Hy - Gz = 0, 

(11.310) { -Ex + (B - 7f)2/ - Fz = 0, 

( -Ox -Fy+(C - K)z = 0. 

Since these equations have a solution other than # = ^ = 3 = 0, 
it follows that 



(11.311) 



A - K - H - G 
- H B - K -F 
-G -^ C ~ 



0. 



This is a cubic equation for K, and, as we have seen, its three 
roots are the three principal moments of inertia. To sum up: 



320 MECHANICS IN SPACE [SEC. 11.3 

Starting with general axes Oxyz with moments and products of 
inertia A, B, C, F, G, H, the three principal moments of inertia at 
are the values of K satisfying the cubic determinantal equation 
(11.311), and the directions of the three principal axes are given by 
the ratios x:y:z determined by (11.310) when the above values of K 
are substituted. 

The problem of finding principal axes and moments of inertia 
is essentially the same as the geometrical problem of finding the 
directions and magnitudes of the principal axes of an ellipsoid 
from its general equation.* The use of the equations (11.309) 
is most naturally suggested by the problem t of finding station- 
ary values (including maximum and minimum values) of the 
expression 

Ax* + By* + Cz* - 2Fyz - 2Gzx - 2IIxy, 
subject to the condition x 2 + y 2 + z 2 = 1. 

Method of symmetry. 

For a body which exhibits symmetry, it is often possible to 
find principal axes of inertia very simply. 

In Sec. 3.1 the idea of symmetry was used in connection with 
mass centers. A more thorough discussion requires the concept 
of a covering operation, which we now proceed to define. 

If we rotate a body of revolution about its axis through any 
angle, we do not alter the distribution of matter the whole 
body appears exactly as before. Similarly, if we turn a thrce- 
bladed propeller about its axis through an angle 2ir/3, the final 
distribution of matter is that with which we started. These 
rotations are examples of covering operations. In general, a 
covering operation for a body is a transformation which does not 
alter the distribution of matter as a whole, although the individual 
particles are moved. In the case of a curve or surface, where no 
distribution of matter is involved, a covering operation is a 
transformation which leaves the curve or surface unchanged as 
a whole. The covering operations which we shall consider are 
(i) a rotation about a line or axis, and (ii) a reflection in a plane. 

* Cf. D. M. Y. Sommerville, Analytical Geometry of Three Dimensions 
(Cambridge University Press, 1934), Chap. VIII. 

t Cf. R. Courant, Differential and Integral Calculus (Blackie & Sono, 
Ltd., Glasgow, 1936), Vol. II, pp. 18&-191. 



SEC. 11.3] KINEMATICS 321 

Whenever there exists a covering operation* for a body, the 
body is said to possess symmetry. If the covering operation is a 
rotation through an angle 2ir/n about an axis (where n is a 
positive integer other than unity), this axis is called an axis of 
n-gonal symmetry; for n = 2, 3, 4 the symmetry is digonal, 
trigonal, tetragonal, respectively. Thus the axis of a three- 
bladed propeller is an axis of trigonal symmetry; for a two- 
bladed propeller the axis is of digonal symmetry. If the covering 
operation is a reflection in a plane, then that plane is a plane of 
symmetry for the body. 

When we speak of an axis of symmetry, without qualification, 
we understand that a rotation through any arbitrary angle is a 
covering operation. A surface of revolution has this type of 
symmetry. 

We now return to the problem of finding principal axes of 
inertia for a body possessing symmetry. In this connection 
we have the following theorem: A covering operation for a body, 
which leaves a point of the body unchanged, is a covering operation 
for the momental ellipsoid at 0. The proof of this theorem 
depends on the following facts, which hold for any distribution 
of matter whether symmetrical or not and are easily proved: 

(i) when a body is rotated about a line, the momental ellipsoid 
at any point on the line turns with the body; 

(ii) when a body is reflected in a plane, the momental ellipsoid 
at any point on the plane is also reflected in this plane. 

When the rotation (or reflection) is a covering operation for the 
body, the distribution of matter is unaltered, and the momental 
ellipsoid at a point on the axis of rotation (or in the plane of 
reflection) is the same as before. The rotation (or reflection) 
is therefore a covering operation for the momental ellipsoid 
also, and so the theorem is proved. 

Now we know from the geometry of the ellipsoid that, when the 
axes are unequal, there are only very special covering operations; 
these are (i) a rotation through an angle TT about a principal axis 
and (ii) a reflection in a principal plane. If the ellipsoid has 
more general covering operations, it must necessarily be of 
revolution or, in particular, a sphere. Thus, for example, if a 
rotation through an angle 2^/3 is a covering operation, the 

* Other than a rotation through four right angles; this is a trivial opera- 
tion, since it leaves every particle of the body back in its original position. 



322 MECHANICS IN SPACE [Sue. 11.3 

ellipsoid must be of revolution. If a rotation through an angle ir 
about a line L is a covering operation, then L must be a prin- 
cipal axis. If a reflection in a plane n is a covering operation, 
then II must be a principal plane. 

We shall now apply these facts to the momental ellipsoid. 
The truth of the following statements will be obvious: 

(i) An axis of n-gonal symmetry is a principal axis of inertia 
at any point of itself. (Example: a two-bladed propeller.) 

(ii) At any point on an axis of trigonal or tetragonal sym- 
metry, the momental ellipsoid has this axis for axis of revolution, 
and two of the principal moments of inertia are equal. (Exam- 
ple: a three- or four-bladed propeller.) 

(iii) The normal to a plane of symmetry is a principal axis 
of inertia at the point where it cuts the plane of symmetry. 
(Example: the hull of a ship.) 

Principal axes of inertia for a number of bodies are given in 
the table on page 324. In each case an argument, based on the 
ideas of symmetry, can be used to verify that the principal axes 
are given correctly. 

The momental ellipse. 

Let us now consider a distribution of matter in a plane n, 
and let Ox, Oy be rectangular axes in this plane. Since H is 
a plane of symmetry, its normal at is a principal axis of inertia, 
and the section of the momental ellipsoid at by the plane II 
is a principal section; it is called the momental ellipse at O. 

If A and B denote the moments of inertia about Ox, Oy, 
respectively, and H denotes the product of inertia with respect 
to planes through Ox, Oy, perpendicular to n, the equation of 
this ellipse is 

(11.312) Ax 2 - 2Hxy + By 2 = 1. 

(We see this by introducing the third axis Oz and putting 2 = 
in the equation of the momental ellipsoid.) It is clear that the 
principal axes of this ellipse are principal axes of inertia at 0. 
To find them we proceed as follows. 

Let Ox', Oy' be new axes at 0, Ox' making an angle with 
Ox. If (x f ' , y'\ (x, y) denote the coordinates of a point referred 
to the axes Ox'y', Oxy, respectively, then 

x = x' cos y' sin 6, y = x' sin + y f cos 0. 



SEC. 11.3] KINEMATICS 323 

The equation of the ellipse (11.312) referred to the axes Ox', 
(Vis 

A(x' cos B - y f sin 0) 2 

2H(x' cos y' sin 0)(x' sin + y' cos 6) 

+ B(x' sin B + y' cos 0) 2 = 1, 
or, equivalently, 

(11.313) AV 2 - ZH'x'y' + B'y'* = 1, 
where 

A' = A cos 2 - 2// sin cos 6 + B sin 2 0, 
//' = (A - J3) sin cos + # (cos 2 - sin 2 0), 
B' = A sin 2 6 + 2H sin cos + B cos 2 0. 

Now (11.313) represents the equation of an ellipse referred to 
principal axes at its center if H f = 0. Hence, Ox', Oy r are 
principal axes of inertia at if 

2/7 

(11.314) tan 20 = _ * 



The two values of in the range (0, TT) satisfying this equation 
give the directions of the two principal axes. The complete set 
of principal axes at are Ox', Oy f , and a line perpendicular to 
them. % 

This method of finding principal axes of inertia at a point can 
be applied to any case where one principal axis tE^vthe point is 
known; it need not be restricted, as here, to the case of a plane 
distribution of matter. In particular, it applies to any body with 
a plane of symmetry or an axis of digonal symmetry. 

Moments of inertia of some simple bodies. 

The table on the following page gives the principal axes and 
moments of inertia at the mass center for some simple bodies. 
The moments of inertia about the axes Ox, Oy, Oz are denoted 
(as usual) by A, B, C, respectively. In all cases the bodies are 
homogeneous, i.e., of constant density. 

Some of the moments of inertia given in the table have already 
been calculated in Sec. 7.1. We shall give the calculations for 
the ellipsoid and leave the reader to verify the others for himself. 

The equation of an ellipsoid E with semiaxes a, b, c, referred 
to principal axes at its center, is 



324 



MECHANICS IN SPACE 



[SEC. 11.3 



The moment of inertia about the #-axis is given by 
A CCC (V* + 2 2 ) dx dy dz, 



Body 

(mass = tn) 


Principal axes 
(at mass center 0) 


Principal moments 
of inertia 


Rectangular plate (edges 
2a, 26). 


Ox, Oy parallel to edges 
2a, 26, respectively; Oz 
perpendicular to plate. 


4 = Imb*, B = \mn\ 
C = fcnfa 1 + & 2 )- 


Solid rectangular cuboid 
(edges 2a, 2fe, 2c). 


Ox, Oy, Oz parallel to 
edges 2a, 2b, 2c, respec- 
tively. 


A = im(6 2 + c 2 ), 
B = \m(c* 4- a 2 ), 
C = imCa 2 + & 2 ). 


Circular plate (radius a) . 


Ox, Oy in plane of plate; 
Oz perpendicular to 
plate. 


A = K = ima 2 , 
C = i/na. 


Elliptical plate (semi- 
axes a, &). 


Ox, Oy along semiaxes a, 
b, respectively; Oz per- 
pendicular to plate. 


A - lmb z , B = Iwa 2 , 
C = im(a -h 6 2 ). 


Solid circular cylinder 
(radius a, length 21). 


Ox, Oy perpendicular to 
axis; Oz along axis. 


A - = 
Aw(3 f + 4/ 2 ), 
C = i?wa. 


Solid elliptical cylinder 
(semiaxes a, 6; length 
2i). 


Ox, Oy along semiaxes a, 
6 of section, respec- 
tively; Oz along axis of 
cylinder. 


A = ^wOb 2 + 4i 2 ), 
^ = iV/(3a 2 + 4Z 2 ), 
C Y = JwCo* + & 2 ). 


Sphere (radius a). 


Ox, Oy, Oz any three per- 
pendicular lines. 


A = ^ = C = frrta 2 . 


Solid ellipsoid (semiaxes 
a, 6, c). 


Ox, Oy, Oz along semiaxes 
a, b, c, respectively. 


A = im(6 2 + c 2 ), 
5 = im(c 2 + a 2 ), 
C = U(a 2 + & 2 ). 



where p is the density^ jind the integration extends throughout 
the ellipsoid E. We put x r = x/a, y' = y/>b, z' = 2/c and obtain 



SEC. 11.3] KINEMATICS 325 

^4. err 
P j j j 

where the range of integration is now the interior of a unit 
sphere S. From the symmetry of S, 



fff 



dx ' dyf dz ' = z ' 2 dx ' dyf dz ' 



(S) 

But this last integral has already been calculated in Sec. 7.1; it is 
the moment of inertia of a sphere (of unit radius and density) 
about a diameter and has the value 8?r/15. Hence, 



A = (b* + c*) 

as given in the table. The values for B and C follow in exactly 
the same way. 

The following rule, known as Routh's rule, summarizes most of 
the results given in the table on page 324: For solid bodies of the 
cuboid, elliptical cylindrical, and ellipsoidal types, the moment of 
inertia about a principal axis through the center (and parallel to the 
generators, in the case of the elliptical cylinder) is equal to 

m(a 2 + b 2 ) 



where m is the mass of the body, a, b are the semiaxes perpendicular 
to the principal axis in question, and n = 3, 4, or 5 according 
as the body belongs to the cuboid, elliptical cylindrical, or ellipsoidal 
type. 

The methods of decomposition and differentiation. 

If we wish to calculate a moment of inertia, we can always do 
so by evaluating a multiple integral. But in many cases there 
are simpler methods. One method is to divide the body into a 
number of parts, for each of which the moment of inertia is 
known; by adding the moments of inertia of these parts, we 
obtain the required result. This is the method of decomposition 
and has been used already in Sec. 7.1. 

Another method, known as the method of differentiation, can 



326 ' MECHANICS IN SPACE [SEC. 11.3 

be used to find the moment of inertia of a shell when the cor- 
responding moment of inertia for a similar solid is known. 

As an example, let us find the moment of inertia of a spherical 
shell about a diameter. We first consider a uniform solid sphere 
of density p and radius r. Its moment of inertia about a diameter 
is (87r/15)pr 5 . If the radius of this sphere is increased to r + dr, 
the moment of inertia is increased by 

dl = (87r/3)pr 4 dr-, 

this is the moment of inertia of a spherical shell of radius r, 
thickness dr, and mass 4?rpr 2 dr. Hence the moment of inertia 
of a spherical shell, of radius a and mass m, about a diameter 
is f wa 2 . 

Similarly, by considering the ellipsoid 



_ 

Fa 2 A- 2 6 2 

and increasing k to k 4- dk, we can find the principal moments of 
inertia at the center of a thin shell bounded by two such ellipsoids. 
The reader will have no difficulty in showing that, for A* = 1, the 
results are 

W + c 2 ), im(c 2 + a 2 ), im(a 2 + 6 2 ), 
where ra is the mass of the shell. 

Equimomental systems. 

Two distributions of matter which have the same total mass 
and the same principal moments of inertia at the mass center 
are said to be equimomental systems. For example, a hoop of 
mass m and radius a/V2 is equimomental with a circular plate 
of mass m and radius a. 

Such systems are interesting on account of the following 
fact: Two rigid bodies* which are equimomental have the same 
dynamical behavior. By this we mean that two such bodies, 
when acted on by identical force systems, will behave in the 
same way; if the bodies were fixed inside two identical boxes, 
we should not be able to distinguish between them. This result 
will be evident when we have developed the general principles of 
dynamics in Chap. XII. 



SBC. 11. 4] KINEMATICS 327 

11.4. KINETIC ENERGY 
The kinetic energy of a rigid body with a fixed point. 

Consider a rigid body turning about a fixed point with 
angular velocity <>. A particle P of this body, with velocity 
q and mass 5m, has kinetic energy ^5m g 2 (cf. Sec. 5.1); the 
kinetic energy of the body is 

(11.401) T = iS dm - q*, 

where the summation extends over all particles of the body. 
We seek an alternative expression for T 7 , involving the angular 
velocity <> and the principal moments of inertia at 0. 

Let Oxyz be any rectangular axes at 0, and i, j, k unit vectors 
along them. Resolving vectors in the directions of these axes, 
we write 

r = xi + yj + 2k, w coj + co 2 j + w 3 k, 

> 
where r = OP. For the velocity q of P, we have, by (11.201), 

(11.402) q=wXr 

= (cdaz - w 3 7/)i + (usz - i0)j + (any - a>2)k. 

Hence, by substitution from (11.402) in (11.401), we obtain 

2T = S dm [(ci>23 w 3 ?/) 2 + (ws wiz) 2 + (wi?/ w 2 .r) 2 ] 

= co 2 . 2 dm - (if + z 2 ) + col 2 6m (z 2 + a* 2 ) + co2 6w - (x 2 + ?/) 

2a>2co3S 5m 2/2 2a>3Wi 25m z# 
or 



(11.403) T = i(Aa> 2 + Bui + Cul - 2Fco 2 co 3 - 2(7co 3 a>i 

2//C01C02), 

where -4, #, C, F, (j, H are the moments and products of inertia 
for Oxyz. If the axes are principal axes of inertia, then 

F = G = H = 0, 
and we obtain, as the required expression for the kinetic energy, 

(11.404) T = KAco? + B<*\ + Cw|), 

where A, B, C are now principal moments of inertia. 

The expression (11.404) is valid only when the axes Oxyz 
are principal axes of inertia at 0; for other axes, it is evident 



328 MECHANICS IN SPACE [SEC. 11.4 

from (11.403) that T involves both products and moments of 
inertia. If we use axes with directions fixed in space, not only 
will T involve both products and moments of inertia worse 
still, these will vary with the time. To avoid these complica- 
tions, it is customary to use axes which are permanently principal 
axes of inertia at 0, so that the simple formula (11.404) holds at 
any time and A, J5, C are constants. 

In general the principal axes at are fixed in the* body. But 
if the momental ellipsoid at is of revolution, only one of them 
need be so fixed; the other two may be any perpendicular lines 
in the plane perpendicular to the axis of revolution. It might 
appear that the use of such axes, fixed neither in space nor in the 
body, would introduce a needless complication. But actually 
it simplifies considerably the theory of tops and gyroscopes. 

For a rigid body turning about a fixed line L through 0, it 
is easily seen that the formula (11.403) simplifies to the formula 
(7.116), given in the two-dimensional theory. We have merely 
to take Oz along L; then on = co 2 = 0. co 3 = co, and (11.403) 
gives 

T = iGV, 

where C is the moment of inertia about L. 

The kinetic energy of a rigid body in general. 

Let us now find the kinetic energy T of a rigid body moving 
quite generally in space. Applying the theorem of Konig 
(cf. Sec. 7.1), we have 

(11.405) T = %mql + T', 

where m = mass of body, 

#o = speed of mass center, 

T r = kinetic energy of motion relative to mass center. 
But the mass center may be regarded as a base point in the body; 
and so, as explained in Sec. 11.2, the motion relative to the mass 
center is that of a rigid body turning about a fixed point. Thus, 
T' is given by (11.404) with a proper interpretation of the 
symbols. We therefore have 

(11.406) T = %mql + i(A? + Bu\ + Cco 2 3 ), 



SEC. 11.5] KINEMATICS 329 

where A, B, C = principal moments of inertia at the mass center, 
i, w 2 , ws = components of the angular velocity co in the 
directions of principal axes of inertia at the 
mass center. 

In applying the principle of energy, proved in Sec. 5.2, to 
particular systems, we need expressions for kinetic energy.. 
For a particle the kinetic energy is simply w</ 2 ; for a rigid body, 
we have the formulas (11.404) and (11.406). With the aid of 
these fundamental formulas, we find no difficulty in calculating 
the kinetic energy of any system. 

11.5. ANGULAR MOMENTUM 

The angular momentum of a particle about a line was defined 
in Sec. 5. 1 as the moment of the linear momentum vector about 
the line in question. Now in dealing with moments of vectors 
in three dimensions, it is the vector moment about a point which 
is fundamental, rather than the scalar moment about a line. 
Accordingly, we define the angular momentum of a particle as a 
vector; the scalar angular momentum defined in Sec. 5.1 is, 
of course, merely one component of the vector defined here. 

Angular momentum of a particle and of a system of particles. 

Consider a particle of mass m, moving with velocity q relative 
to some frame of reference S. The linear momentum is wq 
(cf. Sec. 5.1). We define the angular momentum h, about any 
point 0, as the moment of raq about 0; hence, by (9.301), 

(11.501) h = r X mq, 

where r is the position vector of the particle relative to 0. It 
is clear that h depends on the frame of reference used in the 
measurement of q. 

For a system of particles, the angular momentum is the vector 
sum of the angular momenta of the several particles. Let TW,, 
r t , q t denote the mass, position vector (relative to a point 0), 
and velocity of the zth particle, respectively. The angular 
momentum about is 

n 

(11.502) h = 2) (r< X m t q t ), 

= i 

where n is the number of particles in the system. 



330 MECHANICS IN SPACE [SEC. 11.5 

We note that, if is fixed in the frame of reference, then 
q t = ft. In that case the components of h along rectangular 
axes fixed in the frame are 



(11.503) 



i 



Let us now consider the effect of changing the frame of refer- 
ence. Let S' be a new frame, having a velocity q of translation 
relative to S. Then the velocities q t , q( of a particle relative to 
S, S', respectively, are connected by 

(11.504) q< = q + q5, 

according to (11.108). The angular momenta about are then 

h = i) (r. X m t q t ) for S; h' = Y (r< X m>q' z ) for S'. 
t=i =i 

Substituting from (11.504) in the expression for h, we find 

(11.505) h = (^ m t r t ) X q + h 7 . 



If is the mass center, ^ m r * = 0* an ^ s ^ e fi rs * term on the 



right vanishes. This gives the following remarkable result: 
Angular momentum about the mass center is the same for all frames 
of reference in relative translational motion. Generally it is most 
convenient to use a frame of reference in which the mass center 
is fixed. 

In speaking of angular momentum about a point 0, we shall in 
future always understand a frame of reference in which is 
fixed. 

Angular momentum of a rigid body. 

The most interesting application of (11.502) is to the case 
of a rigid body turning about 0. The formulas which we are 
about to develop are fundamental in gyroscopic theory. 



SEC. 11.51 KINEMATICS 331 

In a slightly different notation, we have, for the angular 
momentum about 0, 

(11.506) h = S(r X 8m - q), 

where 8m is the mass of a typical particle, r its position vector, 
and q its velocity; the summation extends over all particles in 
the body. But, by (11.201), 

q = <o X r, 

where CD is the angular velocity of the body. Hence, 

(11.507) h = S dm [r X (o X r)] = 2dm- [or 2 - r(co r)]. 

Let us resolve this vector along an orthogonal triad i, j, k 
at 0. In the usual notation, we write 

r = xi + y] + zk, co = wii + w-j + cojs 

and denote by A, B, C, F, G, H the moments and products of 
inertia with respect to the triad i, j, k. The component of h 
in the direction of i is 



hi = 2 5m [o>i(z 2 + 2/ 2 + z 2 ) x( 

= coiS 8m (y 2 + z 2 ) co 2 S 8m xy co 3 S 5m 22 
= A&I 7/OJ2 G&Z. 

Similar expressions for the components 7i 2 and /i 3 are found in 
the same way; the complete set of components is 

hi = 
/i 2 = 



The structure of these formulas should be compared with the 
array (11.305). 

If i, j, k aro principal axes of inertia at 0, then F = G = H = 0, 
and these formulas are greatly simplified. They become 



(11.509) hi = Acoi, h = w 2 , /i 3 = Cw 3 , 
where A, B, C are now principal moments of inertia at 0. 



332 MECHANICS IN SPACE [SEC. 11.6 

As in the case of kinetic energy, it is usual to choose the 
coordinate vectors i, j, k in directions which are permanently 
principal axes of inertia at 0. With such a choice for these 
vectors, the simple formulas (11.509) hold at any time, and 
A , By C are constants. 

If the body is constrained to rotate about a fixed axis, we may 
take k along this axis. Then coi = 0, w 2 = 0, o> 3 = w, and 
(11.508) gives 

(11.510) hi = -G, h* = -Fw, h, = GV 

Thus the angular momentum vector doos not lie along the axis of 
rotation, unless the latter is a principal axis of inertia. However, 
the component /& 3 along the axis of rotation is equal to the product 
of the moment of inertia about that axis and the angular velocity. 
This is in agreement with (7.117). 

11.6. SUMMARY OF KINEMATICS, KINETIC ENERGY, 
AND ANGULAR MOMENTUM 

I. Kinematics of a particle. 

Velocity: 

(11.601) q = -r = i, (i = unit tangent vector). 
Acceleration: 

(11.602) f = -77 = .si + j, (j = unit principal normal 

at p 

vector) . 

II. Kinematics of a rigid body. 

(a) Rigid body with a fixed point: 

Motion described by vector o>; velocity of any particle of the 
body is 

(11.603) q - Q X r. 

(6) Rigid body in general motion: 

Motion described by vectors q^, <>; velocity of any particle of 
the body is 

(11.604) q = q^+co Xr. 



Ex. XI] KINEMATICS 333 

III. Moments and products of inertia. 

(a) General formulas: 



(A = Sra(2/ 2 + z 2 ), B = Sm(z 2 + a: 2 ), 
C = 2 
F = Zroys, Cf = 2wz.r, // = S 

(11.606) / = Aa 2 + B(3* + 6V - 2F(3y - 2Gya - 

(b) Momental ellipsoid (r = \/\/l): 
General form: 

(11.607) Ax* + By* + Cz* - 2Fyz - 2Gzx - 2Hxy = 1. 
Form for principal axes: 

(11.608) Ax* + By 2 + Cz* = 1. 

(A, B, C are principal moments of inertia; F G H = 0.) 

IV. Kinetic energy. 

(a) Particle: 

(11.609) T = %mq*. 

(b) Rigid body with a fixed point (principal axes) : 

(11.610) T = i(Awf + Bu\ + CVO- 

(c) Rigid body in general motion (principal axes at mass 
center) : 



(11.611) T = Imql + i(Ao> 2 + #co 2 + Ca, 2 3 ). 

V. Angular momentum. 

(a) Particle: 

(11.612) h = r X mq. 

(b) Rigid body turning about a point (principal axes) : 

(11.613) h = Awii + J5co 2 j + Co> 3 k. 



EXERCISES XI 

1. What is the kinetic energy of a homogeneous circular cylinder, of 
mass m and radius a, rolling on a plane with linear velocity ? 



334 MECHANICS IN SPACE [Ex. XI 

2. For a certain orthogonal triad of axes at the moments of inertia of a 
body are 3, 4, 5, and the products of inertia vanish. What is the greatest 
moment of inertia of the body about any line through 0? 

3. A rod, of length 2a and mass m, turns about one end 0, describing a 
cone with semivertical angle a. It completes a revolution in time T. Find 
the magnitude and direction of the angular momentum about 0. 

4. A rigid body is turning about a fixed point 0, and Oxyz are rectangular 
axes. If the components of velocity of the particle with coordinates (1, 0, 0) 
are (0, 2, 5), find the component in the direction of the x-axis of the velocity 
of the particle with coordinates (0, 0, 1). 

6. Find the length of a homogeneous solid circular cylinder of radius <z, 
given that the momental ellipsoid at the mass center is a sphere. 

6. A body turns about a fixed point. Prove that the angle between its 
angular velocity vector and its angular momentum vector (about the fixed 
point) is always acute. Show that, if the principal moments of inertia 
Ay B, C are all different, then the angle vanishes only if the body is turning 
about a principal axis. 

7. Find the moment of inertia of a solid homogeneous cube about an 
arbitrary line through its center. 

What are the principal axes of inertia at a corner? 

8. Find the moment of inertia of a rectangular plate 3 ft. by 4 ft., of mass 
20 lb., about a diagonal. 

9. Find the components of velocity and acceleration along the para- 
metric lines of spherical polar coordinates r } 6, <f>, for a particle moving in 
space. 

Check your formulas by applying them to the following special cases: 

(i) <j> = constant; (ii) ?r. 

10. A car drives round a curve of constant curvature at constant speed. 
What is the magnitude and direction of the instantaneous acceleration of the 
highest point of a tire? 

11. A uniform circular disk of radius a and mass m is rigidly mounted on 
one end of a thin light shaft CD, of length 6. The shaft is normal to the 
disk at its center C. The disk rolls on a rough horizontal plane, D being 
fixed in this plane by a smooth universal joint. If the center of the disk 
rotates about the vertical through D with constant angular velocity n, find 
the angular velocity, the kinetic energy, and the angular momentum of the 
disk about D. 

12. A car is turning a corner, the middle point of the back axle describing 
a circle of radius r. If the length of the axle is 2a and the wheels are regarded 
as uniform disks, each of radius b, prove that the ratio of the kinetic energies 
of the back wheels is 

6(r + a) 2 + ft 2 
6(r - a) 3 + &*' 



Ex. XI] KINEMATICS 335 

13. An ellipsoid of revolution with fixed center rolls without slipping on a 
fixed plane. Describe the space and body cones. 

14. A plane is fixed in space. Coordinate axes Oxyz rotate about 0. 
Their angular velocity has components i, w 2 , wa along them. If the equa- 
tion of the plane at any instant is 

Ax + Dy -f Cz = 1, 
prove that 

dA dB dC . R 

-I-- sa J[5a>3 CC02, -T7- = C/C01 /IC03, -yT = /1W2 ~ -DCOl. 

15. A governor consists of two equal spheres, of mass m and radius a. 
They are fixed to the ends of equal light rods, oaoh of length c a, which are 
hinged to a collar on a vortical axle. By means of a light linkage and sliding 
collar, the equality of the inclinations to the vortical of the two rods is 
ensured. If this angle of inclination is arid the angular velocity of the 
governor about its vertical axle is o>, show that the kinetic energy is 



+ co 2 sin 0} + Cw 2 cos 2 0, 
where 

A = m(|a 2 + c 2 ), C = lma z . 

16. Prove that the angular momentum cf a moving system about a point 
O is the sum of the following parts: 

(i) the angular momentum about of a particle moving with the mass 

center and having a mass equal to the total mass of the system; 
(ii) the angular momentum of the system about the mass center. 

17. A steel ball is placed between two horizontal planes, which rotate 
with angular velocities w, ' about vertical axes L, Z/. Assuming that no 
slipping takes placo, show that the center of the ball describes a horizontal 
circle with center in the plane containing L arid L'. Show that the distances 
of this center from L and L' are in the ratio ':w. 

18. For a rigid body in general motion, show that there is no point at rest. 
Show also that, in general, there is one point, and only one, with no 
acceleration. 

19. For a system of particles, prove that the kinetic energy of motion 
relative to the mass center may be expressed in the form 



where m = total mass of system, 

m t = mass of typical particle, 

,- = magnitude of velocity of m^ relative to m,, 
and the summation contains one term for each pair of particles. 



336 MECHANICS IN SPACE [Ex. XT 

20. OA is a light rod of length b which turns with angular velocity ft about 
an axis OB perpendicular to it. A is the middle point of a rod CD of mass m 
and length 2a, hinged to OA at A in such a way that CD is always coplanar 
with OB. If 6 denotes the angle OAC, prove that the components of the 
angular momentum of CD about in the directions OA t OB and a direction 
perpendicular to them are, respectively, 

sin cos 0, w!2(6 2 -f- a 2 cos 2 0), Jma 2 0. 



CHAPTER XII 
METHODS OF DYNAMICS IN SPACE 

The following three principles are fundamental in Newtonian 
mechanics: 

(i) the principle of linear momentum, 

(ii) the principle of angular momentum, 

(iii) the principle of energy. 

General forms for the first and last of these principles have already 
been given in Chap. V ; we shall merely recall them here, stressing 
their applications to dynamics in space. The treatment of the 
principle of angular momentum, given in this chapter, is inde- 
pendent of that given in Chap. V; there is reason for this, since 
in two dimensions angular momentum is a scalar, whereas in 
three dimensions it is a vector. 

We begin our discussion of the above principles by considering 
the simplest of all systems a single particle. 

12.1. MOTION OF A PARTICLE * 
Equations of motion. 

For a particle of mass m acted on by a force P, we have, by 
the fundamental law (1.402), 

(12.101) mf = P, 

where f is the acceleration relative to a Newtonian frame of 
reference. This vector equation can also be written in the form 

(12.102) | (mq) = P, 

where q is the velocity of the particle. In this form, it is often 
referred to as the principle of linear momentum for a particle: 
The rate of change of linear momentum of a particle is equal to the 
applied force. 

By resolving the vectors f and P in the directions of rectangular 
axes OxyZj fixed in the frame of reference, we obtain, as in Sec. 
5.1, the equations 

(12.103) mx - X, my - 7, m& = Z, 

337 



338 MECHANICS IN SPACE [Sic. 12.1 

where X, F, Z are the components of P along the axes. These 
are the equations of motion of a particle in rectangular Cartesians; 
other forms of the equations of motion are obtained below. 

Let i, j, k be unit vectors along the tangent, principal normal 
and binormal to the path of the particle. By (11.103), 



where s denotes arc length along the path and p is the radius of 
curvature. Writing 

P = P! i + P 2 j + P 3 k, 

we obtain from (12.101) the following intrinsic equations of 
motion: 

(12.104) ms = P 1; = P 2 , = P 3 . 

P 

Now let i, j, k be unit vectors in the directions of the para- 
metric lines of cylindrical coordinates (R, <#>, z). By (11.106), 
we have 



f = (& - RWi + i ~ 



Thus, if 



we obtain, as equations of motion in cylindrical coordinates, 
(12.105) m(R - Rp) = P, m ~ ~ (R^) = P , mz = P,. 

Equations (12.103), (12.104), and (12.105) arc probably the 
most useful forms of the equations of motion of a particle. Other 
forms may be obtained by following the same general plan, 
namely, resolution of vectors along a suitably chosen orthogonal 
triad. 

When the path of a particle is to be found, it is better to use 
some form such as (12.103) or (12.105), rather than the intrinsic 
equations (12.104). However, if the path is known beforehand, 
the equations (12.104) are particularly convenient. For exam- 
ple, consider a particle sliding down a smooth curve under 



SBC. 12.1] METHODS OF DYNAMICS IN SPACE 339 

gravity. In this case PI is simply the component of the particle's 
weight in the direction of the tangent and is therefore known. 
Integration of the first equation in (12.104) gives s (and s) in 
terms of the time. The other two equations then give the 
reaction of the curve on the particle without further integration. 
There is a point of interest in connection with (12.104). 
From the last of these equations, it is clear that the path is 
such that the osculating plane contains the applied force. We 
recall that, for a flexible cable in equilibrium [cf. (10.217)], the 
osculating plane also contains the external force; there is a close 
analogy between the two problems. 

Exercise. A particle moves on a smooth surface under no forces except 
the reaction of the surface. Show that its path is a geodesic on the surface. 
Is this result true when the surface is rough? 

Integration of the equations of motion. 

To obtain the equations of motion of a particle is one question, 
but to solve them is another. The second task is much harder 
than the first. Indeed, we may say that only a very few of all 
possible problems in dynamics can be completely solved, if by 
solution we mean the expression of the coordinates as easily 
calculable functions of the time t. However, it is always possible 
to obtain solutions in the form of power series in t. Considera- 
tion of this process leads to the following important general 
theorem: The motion of a particle is determined when its initial 
position and velocity are given. 

Let us sketch the proof of this theorem in the case of a free 
particle, moving in accordance with the equations (12.103). 
We shall suppose that X, F, Z are given functions of x, y y z and 
perhaps of x, y, z, t, also. (Consider, for example, a projectile 
under the action of gravity and air resistance, as in Sec. 6.2.) 
Then the equations (12.103) give X Q , y Q , Z Q in terms of 

(12.106) X , 7/o, Z C , XQ, t/o, , 

the subscript indicating evaluation at t 0. If we differentiate 
(12.103) and consider the resulting equations at t = 0, we see 
that they give the third derivatives of x, y, z with respect to 
t at t = in terms of the quantities (12.106), since X Q , y Q , ZQ 
have already been found. Proceeding in this way, we can 
determine all derivatives of x, y, z at t = in terms of the quan- 



340 MECHANICS IN SPACE [&BC. 12.1 

titles (12.106). Thus, we have all the coefficients in the following 
Taylor expansions for x, y, z: 

x = x Q + x t 
y = 2/o + Vot 
z = z + tot + \z<p + ' ' . 

The above series provide a formal solution of the equations 
of motion and, as we have seen, this solution depends only on 
o, 2/o, z , #o, #o, 2o- For the completion of the proof, it is necessary 
to discuss the convergence of the series; this belongs to the 
theory of differential equations, and we shall merely remark 
that the conditions of convergence (for some range of values 
for /) are satisfied in all the problems we shall consider. 

In order that a solution for the motion of a free particle (not 
necessarily expressed in power series) may be made to fit the 
stated initial conditions, there must be available six constants 
of integration. 

For a set of particles moving under forces depending on their 
positions and velocities, it may be shown by an argument 
similar to that given above that the motion is determined when 
the initial positions and velocities are given. (This is most 
easily seen by means of Lagrange's equations; cf. Chap. XV.) 
The number of constants of integration is double the number of 
degrees of freedom. 

If we regard the universe as composed of particles, this result 
leads to a rather surprising conclusion. If we knew at the 
present moment the position and velocity of every particle in 
the universe and could solve the differential equations of motion, 
we should be able to predict the whole future of the universe. 
Even more surprising, since the motions of all the particles 
could be followed backward in time as well as forward, we should 
be in a position to uncover the history of the universe from its 
beginning. 

Is this practical science? It is not, for such a complete knowl- 
edge of present conditions is quite beyond our power. From a 
philosophical point of view, however, the question is of interest 
the question as to whether the past and future are determined 
by the present. That they are so determined is implied in 
Newtonian mechanics, and it is here that quantum mechanics 



SEC. 12.1] METHODS OF DYNAMICS IN SPACE 341 

introduces a new and revolutionary idea: Nothing is certain, 
only probable. 

Principle of angular momentum. 

By (11.501), the angular momentum of a particle about a 
fixed point is 

(12.107) h = r X wq. 

Let us calculate the rate of change of h. Differentiating (12.107), 
we find 

(12.108) h = fXwq + rXmq 

= q X wq + r X raf 
= r XP, 

where P is the force acting on the particle. In words, the rate 
of change of angular momentum of a particle about a fixed point 
is equal to the moment of the applied force about that point. 

If we take as origin of rectangular axes Oxyz and resolve 
vectors in the directions of these axes, we obtain 

!m(yz zy) = yZ zY, 
m(zx - xz) = zX - xZ, 
m(xy - yx) = xY - yX, 

where X, Y, Z are the components of P. The last of these 
equations is the same as that obtained in Sec. 5.1 for a particle 
moving in the plane Oxy, moments being taken about (or Oz). 

Principle of energy. 

For a moving particle we have, as in Sec. 5.1, 

(12.110) f = W, 

where t is the rate of increase of the kinetic energy and W is 
the rate at which the applied forces do work. This general form 
of the principle of energy is of little use, except in the case where 
the working forces are conservative. Then W = F, where V 
is the potential energy of the particle and (12.110) gives, on 
integration, 

(12.111) T + V = E, 
where E is a constant, the total energy. 



342 MECHANICS IN SPACE [SEC. 

The reader may ask: Seeing that the equations of mot 
(12.103) are three equations for three unknowns (and therei 
mathematically complete), why do we trouble to develop f 
more equations (12.109) and (12.111)? The answer is: r . 
latter equations often give directly pieces of information wh 
can be used in conjunction with (12.103) to simplify the worl 

Example. Let us consider the motion of a particle on a. smooth sph 
We use cylindrical coordinates (72, <, z) with origin at the center of 
sphere, the axis of z being directed vertically upward. Then the equa 
of the sphere is 

(12.112) R 2 = a 2 - z*. 

The forces acting on the particle are its weight mg and the normal reac 
N of the sphere. By resolving these forces along the parametric line 
R, <, and z, we find PR, P<t,, and P e in (12.105); these equations, togel 
with (12.112), are four equations from which we can find N, R, <, z as ft 
tions of the time. This plan of dealing with the motion, though s trail 
forward, is not so simple as that given below. 

We first note that, since N does no work, the principle of energy app 
Now the potential energy of the particle is mgz, and its kinetic energ 
%mq z , where q is the velocity, with components given by (11.105). Hei 
by (12.111), 

(12.113) $m(R* + R*<i>* + z*) -f mgz = mE, 

where E is here used to denote the constant energy per unit mass. Ag 
since N and the weight have no moment about Oz, the angular moment 
about Oz is constant. The components of linear momentum in the R- an 
directions have no moments about Oz; the ^-component is mR$ and 
moment is mR 2 <f>. Hence, 

(12.114) R*<t> = h, 

where h is a constant. This result follows also from the second equal 
of (12.105), since P$ = in this case. When the initial position and veloi 
are known, the constants E and h can be found, and the equations (12.1 
(12.113), and (12.114) provide three equations to determine R, <, and z. 
From (12.112), we have, by differentiation, 



When this value of tt and the value of j> from (12.114) are substitutec 
(12.113), we get 

(12.115) *, = 



SEC. 12.2) METHODS OF DYNAMICS IN SPACE 343 

This is a single equation for z as a function of /; when this equation has been 
solved, (12.112) gives R in terms of t directly, and < can be found by a 
quadrature from (12.114). The solution of (12.115) is given in the next 
chapter. 

12.2. MOTION OF A SYSTEM 
Principle of linear momentum; motion of the mass center. 

We now recall some results established in Sec. 5.2. If m l and q t 
denote the mass and velocity of the zth particle of a system, 
then the linear momentum is 

(12.201) M = J m,q t , 

where n is the number of particles. By (5.206), we have 

(12.202) M: = F, 

where F is the vector sum of the external forces. This is the 
principle of linear momentum in its general form. It may be 
stated as follows: The rate of increase of the linear momentum 
of a system is equal to the vector sum of the external forces. 

If q denotes the velocity of the mass center and m the total 
mass, the linear momentum M is raq and (12.202) gives 

(12.203) wq = F. 

This is the equation of motion for a single particle of mass m 
under a force F, and so we have the following result, already 
stated in Sec. 5.2: The mass center of a system moves like a particle , 
having a mass equal to the mass of the system, acted on by a force 
equal to the vector sum of the external forces acting on the system. 

This alternative statement of the principle of linear momentum 
is particularly useful; it reduces the determination of the motion 
of the mass center of any system under known external forces 
to a problem in particle dynamics. As illustrations, we may 
consider the motion of a high-explosive shell or of the earth in 
its orbit round the sun. To determine the motion of the mass 
center of the shell, we need know only the sum of the forces 
exerted by the air on the elements of its surface and, of course, 
the weight of the shell. Similarly, in the case of the earth, its 
mass center moves like a particle subject to the gravitational 
fields of the sun, moon, and other bodies in the solar system. 



344 MECHANICS IN SPACE [SEC. 1 

Principle of angular momentum. 

By (11.502), the angular momentum of a system of partic 
about a point is 

(12.204) h = J (r t X m t q<). 



Here w = mass of ith particle, 

r t = position vector of zth particle relative to'0, 
q t = velocity of tth particle relative to 0, 
n = number of particles in system. 

In what follows, we shall consider to be either a fixed poi 

in a Newtonian frame of reference or the mass center of t 

system. 

The rate of change of h is 

n ^ 

h = 5) (f t X m^i + r X fn.q<). 

t-i 

Since f t = q t , the first vector product vanishes. Thus we have 
(12.205) h = (r t X 



where f t is the acceleration of the ith particle relative to O. 

If is a fixed point, then f t is acceleration relative to a Ne 
tonian frame, and* so 

iiiA - Pi + P{, 

where P t , P^ are, respectively, the external and internal fore 
on the ith particle. Hence, by (12.205), 

(12.206) h = V r, X P< + V r, X PJ. 



The second summation vanishes, since the internal forces have ] 
moment about any point (cf. Sec. 10.2). Hence, 

(12.207) h = G, 

where G is the total moment of the external forces about tl 
fixed point 0. 



SBC. 12.2] METHODS OF DYNAMICS IN SPACE 345 

If is the mass center, the acceleration of the ith particle 
relative to a Newtonian frame is 

f o + f t , 

where f is the acceleration of relative to S [cf. (11.108)]. 
Hence, the equation of motion of the ith particle is 

m t (f + f t ) = P t + Pi. 
Substitution for m t in (12.205) gives 

(12.208) h = V r X Pi + 2 r t X P' t - ( V m t r t ) X f . 

i=i t = i v i=i ' 

The second summation vanishes as before, and the last vanishes 

n 

since 2) m ^i = 0. Hence, we obtain again an equation of the 

t = i 

form (12.207), where G is now the total moment of the external 
forces about the mass center. 

We may sum up the principle of angular momentum as follows: 
The rate of change of the angular momentum of a system about a 
point, either fixed or moving with the mass center, is equal to the total 
moment of the external forces about that point; in symbols, 

(12.209) h = G. 

The equations (12.203) and (12.209) are fundamental in 
dynamics and, indeed, in statics as well. Like the general 
conditions of equilibrium F = 0, G = 0, they hold for any 
system it may be the whole, or any part, of a given distribution 
of matter. When the system is a single rigid body, (12.203) 
and (12.209) provide two vector equations for q and <o, the 
velocity of a base point (the mass center) and the angular 
velocity of the body (cf. Sec. 11.2). Moreover, when applied 
to a system in equilibrium, for which q and h vanish, they reduce 
to the conditions (10.207), the basic equations in statics. 

Example. As a simple illustration, let us consider a cylinder rolling down 
an inclined plane. The mass center moves in a vertical plane, and so the 
vectors q and F lie in this plane. Resolving them along and perpendicular 
to the inclined plane, we obtain the first two equations in (7.312). The 
angular velocity is parallel to the axis of the cylinder, and the angular 
momentum about the mass center is 



346 MECHANICS IN SPACE [SBC. 12.3 

where / is the moment of inertia about the axis of the cylinder. Since h has 
a fixed direction, (12.209) gives a single scalar equation the third and last 
of the equations (7.312). 

Principle of energy. 

In addition to the principles of linear and angular momentum, 
there is a third general principle the principle of energy. This 
principle, established in Sec. 5.2, is very useful when the working 
forces are conservative. Then it leads to the law of conservation 
of energy, 

(12.210) T + V = E, 

where T and V are the kinetic and potential energies and E is 
the constant total energy. 

The two most common systems in mechanics are the particle 
and the rigid body. For each of these systems, the principle of 
energy is not independent of the principles of linear and angular 
momentum; it may, however, be used in place of any one of the 
scalar equations deduced from (12.203) and (12.209) by resolution 
of vectors. When it is used in this way, the value of the law of 
conservation of energy lies in its simplicity; it involves only 
positions and velocities, not accelerations. 

Exercise. A rigid body turns about a fixed axis with constant kinetic 
energy. Show that the magnitude of the angular momentum h (about a 
point on the axis) is also constant. Is the direction of h necessarily fixed? 

12.3. MOVING FRAMES OF REFERENCE 
In Sec. 5.3 the question was raised: If the laws governing the 
motion of a body in a Newtonian frame of reference are known, 
how does it move when viewed from a frame of reference moving 
relative to the Newtonian frame? This question has been 
answered for a particle moving in a plane; we shall now consider 
three-dimensional motion. 

Frame of reference with translational motion. 

Let S be a Newtonian frame of reference and S' a frame of 
reference which has, relative to S, a motion of translation only. 
For a moving particle, we have, as in (11.108), 

(12.301) f = fo + f, 



SEC. 12.3J METHODS OF DYNAMICS IN SPACE 347 

where fo is the acceleration of S' relative to S. Since S is New- 
tonian, the law of motion is 

mf = P, 

where m is the mass of the particle and P the force acting on it. 
In S' the law of motion is, by (12.301), 

(12.302) mf = P - mf . 

Thus the motion of S' gives rise to the fictitious force wf . 
This means that we can regard S' as Newtonian, provided we 
add to the actual forces a fictitious force mf on each particle. 

Rotating frames; rate of change of a vector. 

Let i, j, k be a triad of unit orthogonal vectors in a frame of 
reference $', which rotates with angular velocity Q relative to a 
Newtonian frame S. Any vector P may be expressed in the 
form 

(12.303) P = Pii + P 2 j + Psk. 

We shall now calculate the rate of change of P as estimated by an 
observer in S. 

In calculating dP/dt, we must remember that not only do 
PI, P 2 , P 3 vary, but also the vectors i, j, k. Straightforward 
differentiation of (12.303) gives 



. 1 . 

Now i is a vector fixed in a rigid body (/S'), which rotates with 
angular velocity ii. We may think of i as the position vector 
of a particle B of this body relative to a base point A, the origin 
of i. Then di/dt is the velocity of B relative to A ; and so, by 
(11.203), di/dt = a X i. The same reasoning applies to j 
and k; thus we have 

(12.305) ^ = O X i, jj = Q X j, * = fl X k. 
Substituting these results in (12.304), we obtain 

(12.306) ?* = ^ + Q X P, 



348 MECHANICS IN SPACE [Sac. 1 

where 

( 12 - 307 > f-ir' + TrJ + Tr* 

We use the symbol 8/8t to denote a partial differentiation 
which i, j, k are held fixed. 

We note that dP/dt consists of two parts. The first pa 
$P/dt, is the rate of change of P as measured by* an obsen 
moving with S'; it may be called the rale of growth, since, 
calculating it, we think of the vector P as changing or growii 
whereas i, j, k remain constant. The second part of dP/dt, vi 
ft X P, is due to the rotation of the triad i, j, k; it may be call 
the rate of transport. Thus, for a rotating frame, the rate 
change of a vector equals rate of growth plus rate of transport. 

Motion of a particle relative to a rotating frame. 

Let S' be a frame of reference which rotates with anguj 
velocity ft about a point O, fixed in a Newtonian frame 
Relative to /S, the velocity q of a moving particle A is, by (12.30< 

(12.308) q^J^I + oxr, 


where r = OA. The acceleration is 

(12.309) f,g_j + axq. 
Substitution from (12.308) gives 

(12.310) f-+ Xr + Ox + O 



Let q' and i' denote, respectively, the velocity and accelerate 
of the particle relative to ', so that 

(12.311) q' - f , f - g- 

Now, 

(dO = SO 50 

(12.312) )dt S<^"*" fit' 

I O X (O X r) = O(O r) - rfi s , 



SBC. 12.3] METHODS OF DYNAMICS IN SPACE 349 

and so (12.310) may be written 

(12.313) f = f + f, + f c , 
where 

(12.314) f t = ^ X r + a(a r) - rJ2 2 , f c = 2Q X q'. 

For a particle fixed in S', q' = 0; then f = and f e = 0, 
so that f reduces to f t . For this reason f t may, in the general 
case, be called the acceleration of transport. The acceleration f c 
is called the complementary acceleration or acceleration of Coriolis. 
We note that the acceleration of Coriolis is perpendicular to 
both Q and q'. 

For a particle of mass m acted on by a force P, the law of 
motion in S is 

mi = P; 

in /S', the law of motion is 

(12.315) mf ' = P - mf t - mf c . 

Thus the rotation of S' gives rise to two fictitious forces, mt t 

and mfc. The last of these is the Coriolis n 

force; the first is intimately related to the 

force usually known as centrifugal force. 

When these two forces are added to the 

actual force P, the law of motion of a 

particle in 8' is precisely the Newtonian 

law we say that S' is reduced to rest by the 

introduction of these fictitious forces. 

Frames with constant angular velocity. 

Let us now consider the case where the 
angular velocity of the rotating frame S' is 
constant. Since Q is a constant vector, it 
determines a fixed axis of rotation through 
0. Let AN be the perpendicular from the 
position of the particle A to this axis (Fig. 
127). Then 



N 




Fio. 127. The vectors 
R and r for a particle A. 



ft r = Qr cos 0, 

where 6 is the angle AON. The acceleration of transport is, 
therefore, 



350 MECHANICS IN SPACE [Sac. 12 

f t = QQr cos 6 - rfi 2 
= ONW - (ON 



where R = NA. In this case the equation (12.315) becomes 

(12.316) mf = P + mRfl 2 - 2mft X q'; 

the fictitious force wRfi 2 is the centrifugal force, as ordinari 
understood. 

For a particle at rest in S', q' = 0, and so the only fon 
required to reduce S' to rest is the centrifugal force. The cond 
tion for relative equilibrium is 

(12.317) P + mRn 2 = 0. 

This is actually the condition used in Sec. 5.3, in discussing tl 
equilibrium of a particle on or near the earth's surface. 

Exercise. Show that in a frame with constant angular velocity j 
reduced to rest, the centrifugal force per unit mass is grad V, whe 
V = -* 



Frames of reference in general motion. 

The above results have been obtained on the supposition th, 
the point O is fixed in a Newtonian frame of reference. If 
moving, the formulas (12.308) and (12.309) for q arid f give mere 
the velocity and acceleration of A relative to 0. The comple 
expressions for velocity and acceleration are obtained by addii 
the velocity qo and the acceleration f of to these expressioi 
for q and f , respectively. 

Thus, relative to a frame S' moving in a general manner, tl 
motion of a particle takes place in accordance with the equatic 

(12.318) mf = P - mf - nd t - mf c , 

where f = acceleration of particle relative to ', 
P = force applied to particle, 
f o = acceleration of base point in ' (relative to a Nei 

tonian frame), 
f, f c = acceleration of transport and acceleration of Corio) 

[cf. (12.314)]. 

For slow-moving frames, for which f and the angular velocity 
are small, the fictitious forces mf , mf t , mf c may not 1 



SEC. 12.4] METHODS OF DYNAMICS IN SPACE 351 

noticeable. However, as remarked in Sec. 5.3, they become 
important for an airplane making a sharp turn or pulling out of a 
power dive; the formula (12.318) enables us to estimate the 
force which interferes with the motion of the pilot's hands in 
manipulating the controls. 

12.4. MOTION OF A RIGID BODY 

The general principles of Sec. 12.2 govern the motion of any 
system. In this section, they are used to find explicit equations 
of motion for a rigid body. 

Rigid body with a fixed point. 

Consider a rigid body constrained to rotate about a fixed 
point 0. By (11.509), the angular momentum about O is 

(12.401) h = Auj. + Bwzj + Cco 3 k, 

where i, j, k = unit vectors in the directions of principal axes 

of inertia at 0, 

A, B, C = principal moments of inertia at 0, 
i, w 2 , o> 3 = components of the angular velocity <o of the 

body in the directions i, j, k. 

As pointed out in Sec. 11.4, in general the principal axes at 
are fixed in the body; in that case the triad i, j, k has the angular 
velocity <a. But if A, B, C are not all different, we may use a 
principal triad which is fixed neither in the body nor in space. 
To allow for all possibilities, we shall denote the angular velocity 
of the triad by Q, noting that & = w if the triad is fixed in the 
body. 
Writing 



we apply (12.306) and obtain 

(12.402) h = + a X h 

+ Cw 3 k + (1 
X 



(Cco 3 



352 MECHANICS IN SPACE [SEC. 12.4 

Now, by (12.209), 

h = G, 

where G is the total moment of the external forces about 0; hence 
(12.402) gives, as the equations of motion of a rigid body with a 
fixed 






where G\ } (r 2 , Gz are the components of G along i, j, k. 

If i, j, k are fixed in the body, so that 2 = o, the equations 
(12.403) become 

!Ai - (B - C)o) 2 o) 3 = Gi, 
#C0 2 - (C - 4)0)30)! = G 2 , 
Co>3 (A #)0)10) 2 = #3. 

These are Euler's equations of motion for a rigid body with a fixed 
point. 

When the working forces are conservative, we can use, in 
place of any one of the three equations in (12.403) or (12.404), 
the following equation, deduced from the principle of energy 
(12.210): 

(12.405) iGlwJ + #o)| + Ccof) + V = E. 

Example 1. A rectangular plate 
spins with constant angular velocity w 
about a diagonal. Find the couple 
which must act on the plate in order 
to tnaintain this motion. 

In Fig. 128, O is the mass center 
of the plate and i, j, k are unit 
vectors along the principal axes of 
inertia at 0; k is normal to the plate, 
and i and j lie in its plane, i being 
parallel to the length. The princi- 
pal moments of inertia at are 




FIG. 128. A rectangular plate spin- 
ning about a diagonal. 



(12.406) 



B = |ma 2 , C = \m(a* + & 2 ), 



where m is the mass of the plate, 2a its length, and 26 its breadth. 

If a is the angle between i and the axis of rotation, so that tan a 
the angular velocity of the plate is 

<o = co cos a i -f co sin a j. 



6/0, 



SEC. 12.4] METHODS OF DYNAMICS IN SPACE 353 

The components of u in the directions i, j, k are, therefore, 

coi = co COS or, o>2 = co sin a, coj 0. 

Substituting these values of coi, co 2 , co 3 and the values of A, B, C from (12.406) 
in the equations (12.404), we get 

(12.407) Gi = 0, G 2 = 0, G 3 = |w(a 2 - 6 2 )a> 2 sin a cos a 



These are the components of the couple G which must act on the plate; we 
observe that the axis of the couple is normal to the plate and turns with it. 
If we suppose the plato to turn in bearings at the ends of the fixed diagonal 
and to be subject only to the reactions at these bearings, then clearly it is 
these reactions which supply the couple G. p]ach reaction, lies in the plane 
of the plate and is of magnitude 

a 2 - 6 2 



i 



(a 2 + 6 2 )* 

As the plate turns, these reactions turn with it. 

Fluctuating reactions of this sort must be avoided in the case of flywheels 
and rotors. It is not enough to make sure 
that the mass center lies on the axis of 
rotation. The rotating body must be 
balanced so that the axis of rotation is a 
principal axis of inertia. It is left to the 
reader to prove, by means of (12.404), 
that the fluctuating reactions vanish for 
any rotating body if, and only if, this 
condition is satisfied. 

Example 2. A circular disk of radius a 
and mass m is supported on a needle point 
at its center; it is set spinning with angular 
velocity co about a line making an angle a 
with the normal to the disk. Find the angular 
velocity of the disk at any subsequent time. 

In Fig. 129, k is a unit vector normal to 
the disk at the center 0, and i, j are fixed 
in the plane of the disk; we suppose j 
chosen so that the initial angular velocity lies in the plane of k and j. 
The angular velocity of the disk at any time is 

<0 = C0]l ~\" COgl "{" CO ski 

at t = 0, 

o>i =0, coz = too sin or, cos coo COS a. , 

The principal moments of inertia at are 

A = B \ma*. C $ma*. 




FIG. 129. Disk spinning 
about its center , i and j fixed in 
the disk. 



354 



MECHANICS IN SPACE 



[SBC. 12.4 



Since the external forces (the reaction at and the weight of the disk) 
have no moment about 0, the equations (12.404) give 






AOl ~ (A - C)W 2 W 8 = 0, 

Aw 2 - (C - A)wao)i = 0, 

CW 3 = 0. 



From the last of these equations it follows that w 3 is constant; hence, 
w 3 = co cos <*. Multiplying the second equation in (12.408) by i( = \/ 1) 
and adding the result to the first of these equations, we get 

A\ i(C A)UQ cos a = 0, 
where = 01 + tw 2 . Since C 2 A, this equation can be written 

too cos a = 0; 
the general solution is 



where is a constant. From the initial conditions, we have 

o = *o sin a, 
and hence 
(12.409) o>i = wo sin a sin(w < cos a), 



w 2 = wo sin cos(w < cos a), 



03 = wo COS a. 



This is the solution of the problem as stated, but it docs not tell us at once how 

the disk moves in space. To find this, we 
must either introduce the Eulerian angles 
defining the positions of i, j, k relative to 
fixed axes, or use a different method. 

Example 3. Find the motion in space of 
the dish considered in Example 2. 

In Fig. 1 30, h is the angular momentum 
vector, o) the angular velocity vector, and 
k a unit vector normal to the disk as 
before; i and j are unit vectors in the 
plane of the disk, but not fixed in it. The 
vector j is taken in the plane determined 
by h and k. We note the following 
j facts: 

(i) Since the external forces have no 
moment about 0, then, by (12.209), h is 
it has a fixed direction 
in space determined by the initial 




FIG, 130. Disk spinning about 

its center; j coplanar with h a constant vecto] 
and k. 

conditions. 

(ii) Since h = A<i)\i -f- ^4w 2 j 4- Cwsk and 
in the plane of j and k. 

(iii) Since the triad i, j, k is not fixed in the disk, its angular velocity a is 
different from u. However, k is fixed both in the triad and in the disk. As 



0, then wi = and <o lies 



SBC. 12.4] METHODS OF DYNAMICS IN SPACE -355 

a point of the triad, the extremity of k has velocity O X k; as a point of the 
disk, it has velocity u X k. Hence, 

(Oil + G 2 j + flak) X k = (wii + 2J + w 3 k) X k, 

and so fti = o>i, 12 2 = w 2 . Since wi - 0, we have fii = 0. 

Applying the general equations (12.403) and making use of the above facts, 
we get 

Ca> 3 w 2 - 0, 
Ad>2 = 0, 

Cd> 3 - 0. 

. Thus, a> 2 , ws, hi, h$ are constants, and S23/fi 2 = fti/wj = Cw 3 /^la>2 = h 3 /hz' f 
the angular velocity Q of the triad has constant magnitude and lies along the 
fixed direction h. The following facts concerning the motion are now 
obvious: 

(i) The disk spins about its normal k at a constant rate wa. 

(ii) The angle between k and h, given by h cos /3 = h s , is constant; the 
normal to the disk moves on a cone with axis h, turning about h at the con- 
stant rate ft. 

(iii) The angle a between <o and k, given by to cos a = ws, is constant; the 
angular velocity vector to describes a cone about the normal to the disk. 
This is the body cone (ef. Sec. 11.2). The angle /3 between o and h is 
also constant, and so the space cone has constant sernivertical angle a 3 
and axis h; it lies inside the body cono. 

The above problem is a special case of the motion of a rigid body with a 
fixed point under no forces, considered in Chap. XIV. 

General motion of a rigid body. 

We now consider a rigid body moving quite generally. Let F 
denote the total external force and G the total moment of the 
external forces about the mass center. By (12.203), the accelera- 
tion f of the mass center (relative to a Newtonian frame) is 
given by* 

(12.410) mi = F, 

where m is the mass of the body. For the motion relative to the 
mass center we have, by (12.209), 

(12.411) h = G, 

where h is the angular momentum about the mass center. This 
last equation is exactly the same as if the mass center were 
fixed, and so can be treated by the methods given above. 

* For simplicity, we drop the subscripts from q and f o, the velocity and 
acceleration of the mass center. 



356 MECHANICS IN SPACE [SEC. 12.5 

Let us resolve the vectors f, F, 6, G along a principal triad 
i, j, k at the mass center. As before, this triad is supposed to 
be permanently a principal triad. Its angular velocity will be 
denoted by ft ; if the triad is fixed in the body, Q = <a, the angular 
velocity of the body. Now, by (12.306), 



where 

q = ui + v j + wk 

is the velocity of the mass center. Substituting for f in (12.410) 
and noting that (1-2.411) leads to equations of the form (12.403), 
we obtain the following scalar equations of motion: 



m(u vQ 3 + 
m(b wQi 
m(w u&i 
<j)i #0)2^3 + 



(12.412) 



Here the constants A, B, C are the principal moments of inertia 
at the mass center. 

The equations (12.412) are six equations for the components 
of velocity of the mass center and the components of angular 
velocity of the body. For any one of these six equations, we 
can substitute the law of conservation of energy, 

(12.413) T + V = E, 

provided the external forces are conservative. In a more explicit 
form, (12.413) reads 

(12.414) im(w 2 + v 2 + w 2 ) + |G4 o>? + B<*\ + CwJ) + V = E. 



Exercise. From the basic equations (12.410) and (12411), deduce the 
principle of energy for a rigid body in the form 

T = F q -h G <o. 

12.6. IMPULSIVE MOTION 

The principles of dynamics, thus far considered in this chapter, 
deal with ordinary or continuous motion. By this we mean 
that the forces acting, and the accelerations produced, are finite. 



SBC. 12.5] METHODS OF DYNAMICS IN SPACE 357 

Sometimes we have to deal with problems in which sudden 
changes in velocity occur. For two-dimensional problems of 
this type, we use the methods of Chap. VIII; similar methods 
for motion in three dimensions will now be developed. 

General equations of impulsive motion. 

Integration of the equations (12.203) and (12.209) from time 
t to time t\ gives* 

(12.501) A(roq) = f'Fctt, Ah = P 1 G dt, 

Jt$ I/to 

where A denotes an increment in the time interval t\ to. These 
equations express the principles of linear and angular momentum 
in integrated form. In words, they read as follows: 

(i) the increment in the linear momentum of a system is equal 
to the total impulse of the external forces; 

(ii) the increment in the angular momentum about a point O 
(either a fixed point in a Newtonian frame or the mass center of 
the system in question) is equal to the time integral of the total 
moment about of the external forces, i.e., the total angular 
impulse about 0. 

In this form the principles of linear and angular momentum 
can easily be applied to problems where sudden changes in 
velocity occur. The method of procedure is essentially that 
given in Chap. VIII, and so we shall give only a brief outline 
here. 

The very short time interval t\ t Q in which the changes 
occur is regarded an infinitesimal. Any finite force will then 
contribute nothing to the total impulsive force 

F = lim f tl F dt. 

<1-><0 J** 

On the other hand, a force P, for which 
P = lim r P dt 



is finite, contributes the impulsive force P to F. If r denotes 
the position vector of the point of application of such a force P 

* For simplicity, we drop the subscript from q , the velocity of the mass 
center. 



358 MECHANICS IN SPACE (SEC. 12.5 

(the position vector being relative to the point about which 
the angular momentum is calculated), then r does not change 
by a finite amount in the infinitesimal time t\ ta. Hence, 

lim f* (r X P) dt = r X lim JP 1 P dt = r X P, 

and the force P contributes the impulsive moment r % X P to the 
total impulsive moment 

6 = lim J o " G dt. 

Then, from (12.501), we have 

(12.502) A(mq) = F, Ah = G, 

where F = total impulsive force = vector sum of external 

impulsive forces, 
G = total impulsive moment = total moment of external 

impulsive forces. 
These are the general equations cf impulsive motion. 

For a rigid body in general motion, the first of the above 
equations gives the change in the velocity of the mass center; the 
second gives the change in the angular momentum (and hence 
the change in angular velocity) about the mass center. For a 
rigid body with a fixed point, the second equation in (12.502) 
alone suffices to determine the change in angular velocity. 

Example. A square plate, of mass m and edge 2a, is suspended from one 
corner 0. It is struck at a corner in a horizontal direction perpendicular to 
the plane of the plate. About what line does the plate begin to turn? 

Let i, j, k be the principal triad of inertia at (Fig. 131); i points upward 
along the diagonal through 0, and j is a horizontal vector in the plane of the 
plate. 

Before the plate is struck, the angular momentum about is zero; imme- 
diately afterward, it is 



h = 4oni -f 

where on, cos, ws arc the components of angular velocity and A, B, C the 
principal moments of inertia at 0. If P is the magnitude of the blow, tho 
external impulsive forces are Pk at the point 

r - -o V2 (i + J), 



SBC. 12.5] METHODS OF DYNAMICS IN SPACE 359 

no moment about 0, the 



and an impulsive reaction Q at 0. Since 
second of the equations (12.502) gives 



+ Bu z j + Co>,k - -o V2 (i + j) X 



Hence, 



3 - 0; 



the plate begins to turn about a line 
in its plane passing through 0. The 
angle 0, between i and this axis of 
rotation, is given by 



__ 2 _ A 

tan & ~~* ~~~ ~~ "75* 

Wl /> 



Since 



B = 

we find 



ra 2 + 2mo 2 = 
tan 0= -J; 




the axis of rotation is indicated in 
Fig. 131 by the vectors. 

The impulsive reaction Q at is FlQ , m.Square pkte, suspended 
easily found. The velocity of the frorn o and struck by a blow p k> 
mass center, immediately after the 
plate has been hit, is 



Thus, from the first of (12.502), we get 

-k =Pk + 0, 



and so 



= -* 



360 MECHANICS IN SPACE [SEC. 12.6 

12.6. SUMMARY OF METHODS OF DYNAMICS IN SPACE 

I. Motion of a particle. 

(a) Equations of motion: 

(12.601) wf = P (vector form) ; 

(12.602) mx = X, my = Y, mz = Z 

(Cartesian coordinates) ; 

(12.603) ms = Pi, m - = P 2 , = P 3 

P 

(intrinsic equations). 
(6) Principle of angular momentum: 

(12.604) h = r X P. 
(c) Principle of energy: 

(12.605) f = TF, (T 7 = w<7 2 , Tf = work done); 

(12.606) T + F = E (conservation of energy). 

II. Motion of a system. 

(a) Principle of linear momentum: 

(12.607) Hi! = F, (M = 2) m t q t ); 

v 1 = 1 7 

(12.608) mq = F (motion of mass center). 
(6) Principle of angular momentum: 

(12.609) h = G (fixed point or mass center), 
(c) Principle of energy: 

(12.610) t = F; 

(12.611) T + V = E (conservation of energy). 

III. Motion of a rigid body. 

(a) Rigid body with a fixed point: 

(12.612) 6 = ~ + QXh = G; 

ot 

SAui (B C)w 2 w3 = Gi, 
Bws - (C - 4)ttjtti = (J 2 , 
(7d>3 (A J5)cOiOJ2 = CrsJ 

(12.614) i(Af + Bwi + CJ) + V = E 

(conservative forces). 



Ex. XII] METHODS OF DYNAMICS IN SPACE 361 

(6) Rigid body in general: 

( mf = F (motion of mass center), 
\ i = G (motion relative to mass center) ; 
(12.616) im? 2 + i(A! + Bu\ + C!) + V = E 

(conservative forces). 

IV. Rotating frame of reference. 

Rate of change of any vector: 

(12 ' 617) f - Tt + Q x P " 

V. Impulsive motion. 
General equations: 

(12.618) A(mq) = , Ah =i. 

EXERCISES XII 

1. A heavy particle moves on a smooth surface. Show that its speed is 
the same whenever its path cuts a given horizontal curve on the surface. 

2. Show directly from Euler's equations (12.404) that, if G = and 
A = B, then o> is constant. 

3. A particle is attracted toward a fixed line by a force, perpendicular to 
the line and varying as the distance from the line. Show that its path is a 
curve traced on an elliptical cylinder. 

4. A solid of revolution rotates with constant angular velocity w about a 
fixed axis which passes through its mass center and is inclined to the axis of 
symmetry at an angle a. Prove that the reactions of the axis on the solid 
are equipollent to a couple of magnitude 

(C 4)cu 2 sin a cos a, 

where C is the moment of inertia about the axis of symmetry and A the 
other principal moment of inertia at the mass center. 

6. Explain how a man, standing on a smooth sheet of ice, can turn 
round by moving his arms. 

6. A bar of length 2o is fitted at its middle point with a nut which moves 
without friction on a fixed vertical screw of pitch p; the bar remains hori- 
zontal and turns with the nut. Find the acceleration of the nut. 

7. Two particles, of masses m, w', attract one another according to the 
inverse square law. At time t = 0, m is at the origin and has a velocity u 
along the x-axis, and m' is at the point (a, 6, c) and has velocity components 
(u'j v'j w f ). Determine 

(i) the coordinates of the mass center at time t, 

(ii) the constant areal velocity of the motion of m' relative to m, 

(iii) the constant areal velocity of the motion of m relative to m'. 

8. Two men support a uniform pole of mass m and length 2a in a hori- 
zontal position. They wish to change ends without changing their positions 



362 MECHANICS IN SPACE [Ex. XII 

on the ground, by throwing the pole into the air and catching it. If the pole 
is to remain horizontal throughout its flight and the magnitude of the impul- 
sive force applied by each man is to be a minimum, find the magnitudes and 
directions of the impulsive forces. 

9. An equilateral triangle is formed of three rods, each of mass m and 
length 2a. It hangs from one vertex, about which it is free to turn. A blow 
P is struck on one of the lower vertices in a direction perpendicular to the 
plane of the triangle. Prove that the impulsive reaction on the point of 
support has a magnitude If*. 

10. A solid homogeneous ellipsoid of mass m and semiaxes a, b, c spins 
with constant angular velocity o> about an axis which is fixed in space and 
makes constant angles a, 0, y with the axes of the ellipsoid. Show that the 
components (along the axes of the ellipsoid) of the couple that must act on 
it in order to maintain this motion are 

|ra<o 2 (& 2 c 2 ) cos cos 7 

and two similar expressions. 

11. A particle of mass m moves in a plane under the action of two forces. 
One force is an attraction mk*r toward the origin ; the other is perpendicular 
to the velocity q and has magnitude mk'q. Show that the motion is given 
by an equation of the form 

x + iy - e^ k/t (Ae ict + Be~ ict ). 

How many arbitrary constants (to fit initial conditions) are present in this 
solution? 

12. An insect runs with constant relative speed v round the rim of a wheel 
of radius a which rolls along a straight road with uniform velocity V. Find 
the magnitude and direction of (i) the acceleration relative to the wheel, 
(ii) the acceleration of transport, and (iii) the Coriolis acceleration. Indi- 
cate these accelerations in a diagram. 

13. A free rigid body is at rest. Find three linear scalar equations to 
determine the components of an impulsive force which, applied at an 
assigned point of the body, imparts to that point an assigned velocity. 
Solve these equations in the case where the assigned point lies on one of the 
principal axes of inertia at the mass center. 

14. A thin rod of mass m and length 2a is made to rotate with constant 
angular velocity w about an axis which passes through one end of the rod 
and cuts it at a constant angle a. Reduce the force system exerted by the 
axis on the rod to a force at the fixed end and a couple. 

16. A rigid triangular target is fixed at the corners, and a bullet is fired 
normally into it. Find the region in which the bullet must strike in order 
that no support may experience an impulsive reaction normal to the target 
greater than half the momentum of the bullet. 

16. A uniform circular disk of mass M and radius a is so mounted that it 
can turn freely about its center, which is fixed. It is spinning with angular 
velocity about the perpendicular to its plane at the center, the plane being 
horizontal. A particle of mass m, falling vertically, hits the disk near the 



Ex. XII] METHODS OF DYNAMICS IN SPACE 363 

edge and adheres to it. Prove that immediately afterward the particle is 
moving in a direction inclined to the horizontal at an angle a, given by 

. m(M + 2m) v 
tan a. = 4 TtffTtf , , ( 
M(M -f- 4m) aw 

where v is the speed of the particle just before impact. 

17. A homogeneous ellipsoid of semiaxes 

a, 6, c, (a > b > c) 

is to be mounted on a horizontal axis L in such a way that it may oscillate 
as a compound pendulum with the smallest possible periodic time. What 
position of L relative to the axes of the ellipsoid should be selected? 

18. A crankshaft of mass m, in the form of a letter S formed out of two 
semicircles, each of radius a, spins with angular velocity <o in bearings at its 
ends. Find the magnitudes of the nvictions exerted on the bearings, and 
show the directions of these reactions in a diagram. 

19. A system is sot in motion by impulsive forces applied to certain pre- 
scribed particles. If P is the external impulsive force on a typical particle, 
q its velocity, and &i an arbitrary infinitesimal displacement consistent 
with the constraints (assumed workless), show that 

S(mq - 5r) = S(P 6r), 

where the summation on the left extends over all particles of the system and 
the summation on the right over the prescribed particles. 

Let T be the kinetic energy of the actual motion and T' that of any other 
motion (q') consistent with the constraints and making q' = q for the 
prescribed particles; prove that T < T' (Kelvin's theorem). 

20. A rhombus ABCD is formed of four uniform rods, each of mass m and 
length 2a, smoothly jointed at the vertices. Prove that if the rhombus is in 
motion in its plane, in such a way that A and C are moving along the diagonal 
AC, the kinetic energy may be expressed in the form 

T = 2m(v - 2aw sin 0) 2 + jJwaV, 

whcro v is the velocity of A, w the angular velocity of AB, and 6 the inclina- 
tion of AC to AB. 

Hence, prove by Kelvin's theorem (see Exercise 19^ that, if the rhombus 
is at rest in the form of a square and is jerked into motion by an impulsive 
force applied at A in the direction AC, then the angular velocity imparted 
to the rods is 

3\/2 v 
10 'a 

where v is the velocity imparted to A. 



CHAPTER XIII 

APPLICATIONS IN DYNAMICS IN SPACE MOTION OF 
A PARTICLE 

13.1. NOTE ON JACOBIAN ELLIPTIC FUNCTIONS 

So far, we have used only the elementary functions, alone or 
in combination polynomial, trigonometrical, exponential, and 
logarithmic. We now find it necessary to introduce the elliptic 
function. 

Definition of a function by means of a differential equation. 

A variable y is said to be a function of x when to values of x 
there correspond values of y. In fact, a function is determined 
by a rule which assigns y when x is given. Usually this rule is a 
formula admitting direct calculation of y (e.g., x 2 , 3 sin 2x), but 
we may also use a differential equation to define a function; we 
must, however, assign initial conditions to make the solution 
unique. 

Consider, for example, the differential equation 



with the conditions y = 0, dy/dx = 1 for x = 0. If we had 
never previously heard of the function sin x y this equation and the 
initial conditions would serve to define it. Another way of 
defining sin x is by the differential equation 

(13.101) feY = 1 - y\ 
with the conditions 

(13.102) y = 0, j| > 0, for x = 0. 

A type of differential equation with periodic solutions. 

In connection with elliptic functions, we have to study the 
differential equation 

364 



SEC. 13.1] MOTION OF A PARTICLE 365 

(13.103) 

where k is a constant such that < k < 1. It is really simpler, 
however, to take a more general point of view and study first the 
differential equation 



where f(y) is a general function. We can find out a surprising 
amount about the solutions of this equation without specifying 
the function f(y). We shall, however, assume that f(y) is 
continuous; that it vanishes for y = a and y = b (a < 6), but its 
derivative does not vanish for either of these values; and finally 
that f(y) > for a < y < b. (The right-hand side of (13.103) 
has these properties, if we take a 1, 6 = 1.) 

Let us take a geometrical point of view, regarding x and y as 
rectangular Cartesian coordinates in a plane. The equation 
(13.104) is then a relation between the slope and the ordinate 
on a curve, and a solution, or integral curve, is a curve for which 
this relation is satisfied. 

A number of statements can be made regarding the integral 
curves of (13.104). These will now be given, followed by their 
proofs. 

(A) If we know an integral curve, then that curve translated 
through any distance parallel to the a*-axis is also an integral 
curve. 

(B) If an integral curve starts in the fundamental strip 
a < y ^ 6, it cannot pass out of that strip. 

(C) Every integral curve in the fundamental strip touches the 
bounding lines y = a, y = b and has at no other point a tangent 
parallel to the z-axis. 

(D) There is one, and only one, integral curve touching a 
bounding line at a given point. * 

(E) All integral curves may be obtained from one integral 
curve by translation parallel to the x-axis. 

(F) An integral curve is symmetric with respect to its normal 
at a point of contact with a bounding line. 

* The bounding lines y = a, y b satisfy (13.104), but we do not regard 
them as integral curves. They are singular solutions. 



366 



MECHANICS IN SPACE 



SEC. 13.1 



(G) The z-distance between successive contacts of an integral 
curve with the bounding lines is 



(13.105) 



(H) Any solution of (13.104), y = <t>(x), is a periodic function 
with period 2P, where P is given by (13.105); this means that 



(13.106) 



2P) = 



for all values of x. 

(I) If y = <f>(x) is any one solution of (13.104), then the general 
solution is y = 4>(x + c), where c is an arbitrary constant. 

Some of these properties are shown in Fig. 132. 

Proofs : 

(A) Neither slope nor ordinate is changed by the translation; 
if the relation (13.104) is satisfied by the curve before translation, 
it will be satisfied after translation. 




y=a 

Fia. 132. General character of a solution of the differential equation (13.104). 

(B) If the curve passed out of the strip, /(?/) would become 
negative and dy/dx imaginary. 

(C) By hypothesis, /(a) = /(&) = 0; hence, dy/dx = on the 
bounding lines. Further, f(y) > for a < y < 6, and so 
dy/dx cannot vanish between the bounding lines. 

(D) Let x = zo, y = a be a point on a bounding line. If we 
invert (13.104), take the square root, and integrate, we get 



(13.107) x - zo - - 



dr, 



x 



SEC. 13.1] MOTION OF A PARTICLE 367 

according as dy/dx < or dy/dx > 0. These two equations 
together give (in the neighborhood of x = x ) the unique integral 
curve satisfying the condition of tangency. 

(E) Let C and C' be any two integral curves. We have to 
show that C' may be made to coincide with C by a translation. 
Let C touch y a at x = XQ. Translate C' until it also touches 
y = a at x = XQ. By (A), it is still an integral curve after 
translation; by (D), it coincides with C. 

(F) The two equations (13.107) give the two parts of an 
integral curve, meeting at a point of tangency with y = a. To 
a given y y there correspond equal values of x XQ, except for 
sign. This establishes the symmetry for the parts of the curve 
running up from y = a to y = b. But there is the same sym- 
metry with respect to the normal at a contact with y b. It is 
not difficult to see that this implies symmetry of the whole curve 
with respect to the normal at any point of contact with a bound- 
ing line. If the part of the curve to the right of such a normal 
is folded over the normal, it will coincide with the part of the 
curve on the left. 

(G) This is obvious from (13.107). 

(H) This follows from (G) and the symmetry of the curve. 

(I) This merely expresses (E) in analytic form. Since the 
differential equation is of the first order, we expect just one 
constant of integration. 

The Jacobian elliptic functions. 
Let us now apply our general results to the differential equation 



(13.108) (jy = (1 - 2/ 2 )(l - *V), (0 < fc < 1). 

The fundamental strip is 1 ^ y ^ 1. We define the Jacobian 
elliptic function sn x to be that solution of (13.108) which satisfies 
the conditions 

(13.109) y = 0, ^ > 0, for x = 0. 

ax 

It is evident that sn x depends on the value of Jfc, which is called 
the modulus of the function, and we may write it sn (x, k) ; but 
it is usual to suppress the explicit dependence on k. (In speaking 
of the function, we call it "ess-en-ex.") . 



368 MECHANICS IN SPACE [SEC. 13J 

From the result (I), stated on page 366, we know that the most 
general solution of (13.108) is 

y = sn (x + c), 

where c is an arbitrary constant. Further, by (H), we know 
that sn x is a periodic function. Thus, 



(13.110) sn (x + IK) = sn x, 
where K is the elliptic integral 

(13.111) K = ^ f 1 %L 



dy 



K is, of course, a function of k. 

By (F) the graph of the function sn x has symmetry with 
respect to each normal at a contact with the lines y = 1. But, 
since the right-hand side of (13.108) is an even function of y, the 
graph has a further symmetry. The equation (13.108) and the 
conditions (13.109) are unchanged when we change x into x 
and y into y, and so the curve is unaltered by a reflection in 
the origin. For the general case, shown in Fig. 132, the whole 
curve can be constructed by means of the symmetry when we 
know a half wave, running from y = a to y = b. In the case of 
sn x, we need merely know the curve from y = to y = 1 or, 
equivalently, from x = to x = K. 

The properties of sn x may be summed up as follows: 

= (1 sn 2 j:)(l k 2 sn 2 x), 

(13.112) (0 <*<!), 

snO = 0, f -T- snzj = 1, 

sn (x + 4K) = sn x. 

We now define other elliptic functions, en x and dn x, by the 
equations 

n 3 1 1 ^ / cn2 x ^ ^ ~~ sn2 X) cn o = i, 

(16.116) \ dn 2 Z = 1 - /b 2 sn 2 x, dn = 1, 



SEC. 13.1] 



MOTION OF A PARTICLE 



369 



with the further condition that the 
functions and their derivatives shall 
be continuous. Since k < 1, dn x is 
always positive. It is clear that 
en x has the period 4K and dn x 
the period 2K. 

If we take the square roots of the 
two sides of the first equation in 
(13.112), we get an ambiguous sign. 
However, for continuity, one sign 
must be taken throughout, and that 
sign is fixed by considering x = 0. 
Thus we find 

(13.114) -j- sn x cnx dn x. 
ax 

Differentiation of (13.113) gives 

d 

sn x -j- sn x 
dx 

= sn x en x dn x, 



d d 

en x -j- en x sn x -p- sn x 
dx dx 



and so 

(13.115) -T- en x snrcdna;. 
Similarly, 

(13.116) -T- dn x = fc 2 snzcnz. 

Just as (13.108) is a generalization 
of (13.101) and reduces to it if k = 0, 
so the elliptic functions are general- 
izations of the trigonometric func- 
tions. In fact, if k 0, we have 



(13.117) 



sn x sin x y 
en x = cos x, 
dn x = 1, 



and (13.114), (13.115) reduce to 
familiar formulas. 




370 MECHANICS IN SPACE [SEC. 13.2 

The theory of elliptic functions is extensive, but this very 
brief presentation contains enough to enable us to solve certain 
dynamical problems. For numerical tables of the functions 
sn x, en x, dn x, see L. M. Milne-Thomson, Die elliptischen 
Funktionen von Jacobi (Verlag Julius Springer, Berlin, 1931). 
Tables of elliptic integrals may also be used to find the elliptic 
functions; cf. J. B. Dale, Five Figure Tables of Mathematical 
Functions (Edward Arnold, London, 1903), or E. Jahnke and 
F. Emde, Tables of Functions (B. G. Teubner, Leipzig, 1938). 
Figure 133 shows graphs of the functions, drawn for k* = 0.7; 
this makes K = 2.07536. 

Exercise. Show that, in the limit k = 1, we get 

sn x = tanh x y en x dn x = scch x. 

13.2. THE SIMPLE PENDULUM 
The motion in terms of elliptic functions. 

We can now give the exact solution for the motion of a simple 
pendulum in terms of elliptic functions.* Let m be the mass 
of the bob and a the length of the pendulum. The equation of 
energy (12.111) gives 

(13.201) \mtffr - mga cos B = E, 

where 6 is the inclination of the string to the downward vertical, 
and E the constant total energy. We shall suppose the motion 
to be oscillatory with amplitude a, so that 6 = for = a. 
Then E = mga cos a, and (13.201) may be written 

(13.202) 6 2 = 2p 2 (cos - cos a) = 4p 2 (sin 2 %a - sin 2 $0), 

where p 2 = g/a. 
Let us define <j> by 

(13.203) sin |0 = sin \a sin , 

so that 

| cos \B 6 = sin \a cos < <. 

* Since the motion of a compound pendulum is identical with that of the 
equivalent simple pendulum (cf. Sec. 7.2), the solution now given applies 
also to the compound pendulum; in (13.202) and the subsequent equations, 
we are to put p* = ga/k*, where a is the distance of the mass center from the 
axis of suspension and k the radius of gyration about that axis. 



SBC. 13.2] MOTION OF A PARTICLE 371 

Multiplying (13.202) by i cos 2 |0, we get 

sin 2 v& cos 2 <t> (j> 2 = p 2 sin 2 %a cos 2 <j> cos 2 0, 
or 

(13.204) tf = P 2 - sin 2 \OL sin 2 0). 
If we multiply this equation by cos 2 < and put 

(13.205) y = sin < = Sm ? , k = sin ia, 

sin Tct 

we get 

(13.206) i/ 2 = p 2 (l - ?/ 2 ) (1 - k*y 2 ). 

Except for the constant p 2 on the right, this has the form of the 
equation (13.108). To get the exact form, we define a new 
independent variable by 

(13.207) x = pt 
and obtain 



(13.208) = (1 - *)(! - *V). 

The general solution of this equation is 

(13.209) y = sn (x + c), 

where c is a constant of integration. Honce, we have the follow- 
ing result: The general oscillatory motion of a simple pendulum, 
with amplitude a, is given by 

(13.210) sin $0 = sin $a ni \p(t - )1, 

where to is a constant cf integration, p 2 = g/a, and the modulus 
of the elliptic function is k = sin -Jar. 

We usually find in dynamical problems that, if sn appears, the 
other elliptic functions en, dn have simple physical meanings. 
From (13.210) we get at once, by (13.113), 

(13.211) cos $0 = dn [p(t - Jo)], 
and differentiation of (13.210) gives 

(13.212) 6 = 2p sin $a en [p(t - <)]. 



372 MECHANICS IN SPACE [SEC. 13.2 

The periodic time. 

As we have seen, the periods of sn x and en x are 4X. Thus, 
by (13.210) and (13.212), the motion repeats itself after a time 
4K'/p, and so the periodic time of the pendulum is 



(13213) r = = - 
P P 



Putting y sin <, we get 

(13.214) r -i "-** 

P J Vl - * 2 si 



Now, 



flT 

Jo s 



1.3 



and so we have the following infinite scries for the periodic time 
of the pendulum: 



For very small amplitude a, we get, as a first approximation, 

a 



agreeing with (6.307), where I was used to denote the length. 
The next approximation is 

(13.216) r . 

It is evident from (13.215) that the periodic time increases 
steadily with the amplitude. 



SEC. 13.3] 



MOTION OF A PARTICLE 



373 



13.3. THE SPHERICAL PENDULUM 

A particle of mass m is attached to a fixed point by a light 
string or rod of length a and oscillates under the action of gravity. 
Since the particle is thus constrained 
to move on a sphere, this system is 
called a spherical pendulum. Under 
special initial conditions, a spherical 
pendulum will move in a vertical plane; 
then the motion is that of a simple 
pendulum, discussed in Sec. 13.2. 

Although we shall be able to deter- 
mine the general motion of a spherical 
pendulum in terms of elliptic functions, 
there are two particular motions which 
can be discussed quite simply. The 
first is a motion in which the particle 
performs small oscillations near the 
lowest point of the sphere, and the 
second is motion in a horizontal circle. 



O 



/ 

FIG. 



134. Spherical pendu- 
lum 



Small oscillations (first approximation). 

Let Oxyz be rectangular axes, O being at the lowest point 
of the sphere and Oz being directed vertically upward (Fig. 134). 
If i, j, k is a unit orthogonal triad along the axes, the position 
vector of the particle is 



(13.301) 



r = xi + yj + zk. 



The force P on the particle is made up of gravity and the tension 
(S) in the string. Now the direction cosines of the string 
(running from the particle to the point of support) are 



x 
a 



l/ ? 
a 



and so 
(13.302) 



P = - 



- [xi + yj + (z - a)k] - 



So far the expressions are exact. But if x and y arc small, z 
is a small quantity of the second order, since the plane 2 = 
touches the sphere. Hence, we have as equation of motion, 



374 MECHANICS IN SPACE [Sue. 13.3 

omitting small quantities of the second order, 

(13.303) m(A + yj) = - ^i - ^ j + (S - m?)k. 
Comparing the coefficients, we see that S = mg, and 

(13.304) x + p*x = 0, if + P 2 */ = 0, U 2 = |Y 

These are simple harmonic equations, as in (6.403). As far as its 
projection on the horizontal plane is concerned, the bob of the 
pendulum moves like a particle attracted toward by a force 
proportional to the distance from 0. As shown in Sec. 6.4, the 
path is an ellipse with center at 0. (When the ellipse degener- 
ates to a straight line, we get the motion of a simple pendulum, 
performing small oscillations.) 

We have idealized the problem by leaving out all consideration 
of frictional resistance. The effect of this is to cause the bob 
of the pendulum to spiral in toward 0, instead of continuing for 
ever in the elliptical path. However, the approximation (neglect 
of z) is perhaps a more serious oversimplification. We shall see 
the effect of this later, when we consider the second approximation. 

The conical pendulum. 

Any prescribed motion of a particle will take place under tile 
action of a suitable force, namely, a force equal to the acceleration 
multiplied by the mass of the particle. Thus the bob of* a 
spherical pendulum may be made to move in any way on the 
sphere defined by the length of the pendulum; to produce this 
motion, it is in general necessary to add a suitable force to the 
weight of the bob and the tension in the string. But if we can 
find a motion in which no such additional force is required, then 
that motion is a possible motion of the pendulum under weight 
and tension alone. 

Consider a motion in a horizontal circle of radius R at constant 
speed q. The acceleration is of constant magnitude q z /R and is 
directed in along the radius of the circular path. Resolving 
along this radius, along the tangent to the circular path, and 
vertically, we find that no additional force is required provided 
that 

S sin = -~; S cos = mg, 



SEC. 13.3) 



MOTION OF A PARTICLE 



375 



where m is the mass of the bob, S the tension, and 6 the inclina- 
tion of the string to the downward vertical (Fig. 135). Since 
sin 6 R/a, elimination of S gives 

gR* 
(13.305) --- 



If 6 denotes the depth of the horizontal circle below the center 
of the sphere, so that b 2 = a 2 R 2 , 
we have 



(13 .306) 



This gives the speed q at which 
a horizontal circle at depth b may (^ 
be described by the bob of the - 

pendulum. When behaving in 
this way, the pendulum is called 




FIG. 135. Conical pendulum. 



a conical pendulum, since the string 
describes a right circular cone. 

Exercise. Find the tension in the string of a conical pendulum moving 
at a depth b. Examine the limits b * a, b 0. 

The general motion of a spherical pendulum. 

To investigate the general motion of a spherical pendulum, we 
take cylindrical coordinates R, <, z, the origin being at the 
center of the sphere and the axis of z directed vertically upward. 
We have already obtained the equations of motion in (12.114) 
and (12.115); they may be written 

(13.307) (a 2 - z 2 )< = h, 

(13.308) z 2 = /(z), 

where 

U3.309, *,.![[<,_.,(.-)_*} 

We recall that a is the length of the string (i.e., the radius of the 
sphere on which the particle moves), and h and E are constants, 
the values of which depend on the initial conditions. We shall, 
for definiteness, assume h positive, so that <t> increases; there is no 
loss of generality here, since we can reverse at will the sense in 
which 6 is measured. 



376 



MECHANICS IN SPACE 



[SEC. 13.3 



Our plan is to solve (13.308) for z as a function of t; then 
(13.307) will give by a quadrature. 

We note that f(z) is a cubic. It is positive for large positive 
values of z; it is negative for z = a; it is positive for values of z 
occurring during the motion, as we see from (13.308). These 
last values must of course lie in the range ( a, a). Since a 
cubic cannot have more than three changes of sign, it follows that 



-a 




Fm. 136. Graph of /(z). 

the graph of /(z) is of the general nature shown in Fig. 136 ; the 
function f(z) has three real zeros, Zi, z 2 , 23, such that 



(13.310) 



a < Zi < Z2 < CL < 



(In exceptional cases, we may have one or more signs of equality 
instead of inequality.) Since /(z) cannot be negative during the 
motion, we see that z oscillates between the values Zi, z 2 . We 
note that the equation (13.308) is of the type (13.104), so that 
all the general results established for (13.104) apply to (13.308). 
To obtain z as a function of t, we proceed as follows: Since 
Zi, 2, 2 3 are the zeros of /(z), we have 



(z - zi)(z - z 2 )(z - 



that z = 2uu. Then (13.308) 



(13.311) /( 

Let us define u 
gives 

(13.312) 

This suggests the differential equation (13.108) for the elliptic 
function sn. Let us define 



jT-j (Z 2 ~ Zi ~ M 2 )(Z 3 Zi W 2 ). 



SEC. 13.3] 



MOTION OF A PARTICLE 



377 



(13.313) 



k = 




Then (13.312) may be written 

(13.314) *> 2 = p 2 (l - v*)(l - 
and so 

(13.315) v = sn [p($ - fe)], 

where Jo is a constant of integration. Hence we have 

!z - Zi = (z 2 - 21) sn 2 [p(t - * )], 
2 2 - z = ( 22 - 20 cn 2 [p(t - to)], 
23 - * = (z 3 - zO dn 2 [p(* - *)]. 

Any one of these three equations gives z as a function of t. We 
note that z has the period z 

9 JT 

(13.317) r = , 

where jf^ is as in (13.111). Since, 
by (13.307), </> increases steadily 
throughout the motion/the path of 
the particle on the sphere is as 
shown in Fig. 137. 

Wo have already seen that the 
particle oscillates between the two 
levels z = Zi and z = z 2 . We 
shall now show that the arithmetic 
mean of those levels lies below the 




FIG. 137. The path of the bob of a 
spherical pendulum. 



center of the sphere, this statement being equivalent to 
(13.318) zi + z 2 < 0. 

We have two different expressions for f(z), (13.309) and 
(13.311) ; they must, of course, be identically equal, and therefore 



(13.319) 



+ 



378 MECHANICS IN SPACE [SBC. 13.3 

From the second of these, we have 

(13.320) 2l + , = - L*!?. 

23 

Since 3i and 22 are each less than a in absolute value and z 3 is 
positive, (13.318) follows at once. 

The path on the sphere is represented analytically by a relation 
connecting z and <j>. Elimination of t from (13.307) and (13.308) 
gives the required relation in the form of a differential equation 






If we look down on the pendulum from a great distance above, 
the bob appears to describe a plane curve, with R and < as polar 
coordinates. The motion resembles that of a particle attracted 
toward a center of force, the areal velocity (^R^<j>) being a con- 
stant for both motions. Just as we considered the apsides of the 
orbit of a particle in a plane, so we can consider the apsides of the 
horizontal projection of the path of the spherical pendulum. 
These points correspond to stationary values of R, i.e., to R = 0. 
Hence z = at these points, since R* + z* = a 2 . As the actual 
path oscillates between the circles at heights z = z\, z = z% on 
the sphere, so the horizontal projection of the path oscillates 
between circles of radii \A 2 z? and \/a z z 2 ,. 

The apsidal angle a is the increment in < corresponding to the 
passage from z = zi to z = z 2 . Thus, by (13.321), 

(13.322) <x = hf" 2 _ d * 

ha m 

(a 2 z 2 ) \/(z Zi)(z z 2 )(z z 3 ) 



Exercise. Find Zi, z 2 , ZB for a conical pendulum with the bob at a depth 6 
below the center of the sphere. 

Small oscillations (second approximation). 

The expression (13.322) is too complicated as it stands to be 
of much interest. But it yields a definite simple result when we 
suppose the oscillations to be small. The reasoning is delicate, 
because, as Zi and Zz tend to a (the lowest point on the sphere), 
the extent of the range of integration tends to zero, and the 



SEC. 13.3] MOTION OF A PARTICLE 379 

integrand tends to infinity. The method of approximation is 
important; the same method may be used in finding the rotation 
of the perihelion of Mercury in the general theory of relativity. 

Before making any approximation, however, we shall first put 
(13.322) into a form in which a, z\, z% are the only constants 
occurring explicitly. To do this, we refer to (13.319). We have 

h 2 / a 2 + 2iZ 2 

^(zi + z 2 + * 3 ), 23 = ~ 






if we eliminate z 3 from these two equations and subtract z from 
the second equation, we obtain 



(13.323) 



where 



23 ~ Z = -r [z(Zi + 2o) + O 2 + ZiZ Z \, 

z\ H- 2 2 



(13.324) S = V(a + zi)(a + z a ), D = V(o ~ *i)(o - 22). 
Substitution in (13.322) gives for the apsidal angle the required 
expression 

(13.325) a - aSD f" F(z)dz, 

j&\. 

where 

F(z] = l 

(a 2 - 2 2 ) Vfe - z) (z - z l )(z(z, + 22) 4- a 2 + i a ] 

We cannot use the binomial theorem to expand negative 
powers of terms which vanish when z, z\, z* tend to a. How- 
ever the term in the square bracket remains finite, and we may 
expand a negative power of it. Putting 

(13.326) z = -a + f, 

where f is small, we get approximately, i.e., neglecting f 2 , 
1 1 



380 MECHANICS IN SPACE [SEC. 13.3 

Hence, we can write 

(13.327) a = aSI - t ^ (z, + z z )J, 

where 

dz 



(a 2 - 



= p 

J*i (a - 



dz 



(13.328) 

./ = r 

(a z) v (z 2 z)(z 

These integrals are evaluated without difficulty by means of the 
substitution 

z = 2 sin 2 + zi cos 2 0, 
and we find 



Substitution in (13.327) gives 

(13.330) = fr Tl + 8D- 

It is evident from (13.324) that S is small; thus the last term 
is small, and we introduce only a negligible error (of the second 
order) if we substitute in the fraction 

D = 2a, Zi + z z = -2a. 
This gives 

(13.331) a = i 

Now if Ri, Rz are the distances from the central vertical to the 
apsides, we have accurately 

2 n 2 __ 2 02 _ 2 _ 2 Of) 7? 7? 

Jtj u- ^j, 7t2 t* ^2, Oxx /ti/t2, 

and so the approximate formula (13.331) becomes 

(13.332) a = 



We saw, in connection with the equations (13.304), that in the 
first approximation the path of the particle is a central ellipse. 



SEC. 13.4] MOTION OF A PARTICLE 381 

For that curve the apsidal angle is far. Now we see, from 
(13.332), that the apsidal angle is a little greater than far. This 
means that the apse advances; the path is approximately a central 
ellipse, but this ellipse turns slowly forward (i.e., in the same sense 
as that in which the particle describes the path). In one rotation 
of the particle, the apse advances through an angle 

(13.333) 4 - 27r = 



4a 2 4a 2 

where A is the area of the ellipse. The advance disappears when 
A = 0, i.e., when the orbit is flattened into the track of a simple 
pendulum. 

This advance of the apse can be shown by fitting a light writing 
device to the bob of the pendulum. This traces the rotating 
elliptical path on a sheet of paper, placed underneath. 

13.4. THE MOTION OF A CHARGED PARTICLE 
IN AN ELECTROMAGNETIC FIELD 

Much of our knowledge of the structure of matter is derived 
from the study of the motion of charged particles (electrons or 
ionized atoms) in electromagnetic fields. Further interest has 
been added to the problem by the invention of the electron 
microscope and other devices, in which streams of electrons 
produce images in much the same way as images are formed by 
rays of light in an optical instrument. 

Electrostatic and magnetostatic fields. 

We shall consider only statical fields, i.e., fields which do not 
change with time. Such fields are produced by electric charges 
at rest in condensers or by steady currents; permanent magnets 
may also be used. 

In an electrostatic field, there exists at each point of space an 
electric vector E. It is the negative of the gradient of an electric 
potential V, so that 

(13.401) E = - grad V. 

The potential V cannot take arbitrary values throughout space; 
it must satisfy Laplace's partial differential equation 



382 MECHANICS IN SPACE [SBC. 13.4 

Similarly, in a magnetostatic field there is at each point of space 
a magnetic vector H, such that 

(13.403) H = - grad Q; 

12 is the magnetic potential, and it also satisfies Laplace 's equation 



Both fields may be present at the same time. The force 
exerted on a particle carrying an electric charge c, moving with 
velocity q, is 

(13.405) P = eE + eq X H, 

if the units are suitably chosen. 

We accept these basic formulas of electromagnetic theory as 
the foundation for our dynamical deductions. 

Before proceeding to discuss special fields, we shall obtain an 
equation of energy from (13.405). If m is the mass of the 
particle, its equation of motion is* 

(13.406) mq = E + eq X H. 

Taking the scalar product of each side with q, we get 

ft (?<q 2 ) = mq q = eE q = -e(grad V) q = - -jg" 

Hence, we have the equation of energy 

(13.407) ira? 2 + eF = constant. 

If the field is purely magnetic, so that V disappears, the speed of 
the particle remains constant. 

Exercise. The potential due to a charge e at the origin is V = e/r, where 
r* = a; 2 -f- y 2 + z*. Verify that this satisfies Laplace's equation, and show 
that the force between two charges at rest satisfies the inverse square law 
(6.502). 

* This is the nonrelativistic equation of motion and is a good approxima- 
tion if the velocity of the particle is small compared with the velocity of light. 
In the accurate relativistic equation, we replace the left-hand side of (13.406) 



SEC. 13.4] MOTION OF A PARTICLE 383 

Motion in a uniform field. 

A simple solution of (13.402) is 

V = ax + by + cz + d, 

where a, 6, c, d are constants. This gives a uniform electric field, 
in which E is a constant vector. Similarly, we may have a 
uniform magnetic field, in which H is a constant vector. 

Let us now suppose that a particle, of mass m and carrying a 
charge , moves in a uniform electric and magnetic field. If r 
is the position vector of the particle, we have as equation of 
motion 

(13.408) mi = cE + ef X H. 

Let us now choose our axes so that Oz is parallel to H. The 
vector equation (13.408) gives the three scalar equations (with 
the usual notation for components) 



(13.409) 



;* + ** 



v -*-- 



m ' 



To complete the solution most conveniently, we introduce the 
complex quantities 

f = x + iy, F = Ei + iE 2 . 

Then the first two equations of (13.409) may be written together 
in the complex form 

(13.410) ^ + 'If ^ * m"" 

This is a differential equation with constant coefficients, and the 
characteristic equation for solutions of the form e nt is 

n ( n + ) = 0. 
\ m / 

Thus the general solution of (13.409) is 
4- iy = f = A 4- 

= C + D< + ! f 



(13.411) 



384 MECHANICS IN SPACE [SEC. 13.4 

where p = eH/m and A, B, C, D are constants of integration; 
A and B are complex, whereas C and Z) are real. These equations 
give the motion of a charged particle in a uniform electric and mag- 
netic field. 

Let us examine this motion in the case where the electric and 
magnetic fields are perpendicular to one another. Then E& = Q 
and the ^-velocity is constant. Let us, for simplicity, assume 
that this component of velocity vanishes and that z = through- 
out the motion. Then the trajectory is described by the complex 
position vector f, as given by the first of (13.411). Let us write 
this equation in the form 

/ 7 -/rA 
(13.412) r - M - jf J = Ber**. 

We recall that any complex number Z may be written in the form 



If Z is a complex position vector, \Z\ is the radius vector, and 
argZ the azimuthal angle. Equating moduli and arguments in 
(13.412), we have 



(13.413) 



[f - (A - )] = 



arg f - A - = tagS - pt. 



If F were zero, the first equation would indicate motion in a 
circle with center A and radius \B\, arid the second equation 
would tell us that the circle is described with constant angular 
velocity p. (The sign shows the sense.) The effect of the 
F-term is simply to impose an additional motion in which the 
center of the circle moves with constant complex velocity 

(13.414) - j = 1 (JB, - iE l ). 

This velocity is of magnitude E/H and is perpendicular to the 
electric vector. 

We sum up our description of the motion of a charged particle 
in perpendicular uniform electric and magnetic fields as follows: 
// started with a velocity perpendicular to H, the particle moves as if 



SEC. 13.4] 



MOTION OF A PARTICLE 



385 



it were attached to the edge of a circular disk which moves in a plane 
perpendicular to H; the disk spins with constant angular velocity 
eH/m, and its center has a constant velocity E/H perpendicular 
to E (Fig. 138). The surprising part of this result is that, on the 
whole, the particle docs not move in the direction of the electric 
field, but perpendicular to it. 




Fio. 138. Motion of a charged paiticle in perpendicular uniform electric and 
magnetic fields. 

Exercise. Suppose the charged particle starts from rest at the origin at 
time t = 0. Starting from (13.412) prove the following facts concerning the 
motion: 

(i) At time t = 2-jr/p, it will be at rest again at a distance 2irE/(pH) from 
the origin. 

(li) If H is very small, and if the particle is allowed to travel for a definite 
finite time ti t then at t = ti its complex position is approximately ^eF/J/m, 
and its complex velocity is approximately eFti/m. 

Motion in a purely electric field and in a purely magnetic field. 

We have worked out (13.411) for the general case in which 
both electric and magnetic fields are present. In the case of a 
purely electric field (H = 0) we return to (13.408), which 



386 MECHANICS IN SPACE [SBC. 13.4 

becomes 

(13.415) mi = E. 

If the field is uniform, the acceleration is constant, and so the 
particle describes a parabolic trajectory like a projectile under 
gravity (cf. Sec. 6.1). The plane of the trajectory is determined 
by the vector E and the initial velocity. 

In the case of a uniform purely magnetic field (# = 0), the 
equations (13.411) read 



r = A + Be- 1 , z = C + Dt, ( p = 

By moving the origin, we can make A C 0; then we have 

(13.416) f = Be~ lpt , z = Dt. 

Hence 

in = i*i, 



D 



_ 
P B 



since these values are constant, it is clear that the trajectory 
is a circular helix, with axis parallel to the magnetic field. The 
azimuthal angular velocity is p = eH/m. 

The simplest motion in a uniform magnetic field is one in 
which the initial velocity is perpendicular to the field. Then 
D = in (13.416), and the trajectory is a circle described with 
constant speed. 

The determination of the charge c and the mass m of an 
electron is a problem of great physical interest. Let us see how 
the results we have established help in that determination. 
The first thing we notice is that and m appear in our equations 
only in the form e/m, and therefore it is only this ratio that we 
can hope to find. It would seem a simple matter to find e/m 
from the circular motion described in the preceding paragraph. 
The angular velocity is eH/m, and so we have, on equating two 
different expressions for the angular velocity, 



. 

R* m 2 ' 

where q is the constant speed and R the radius of the circle. 
(We have squared the two expressions to avoid the question of 
sign, which is of no importance here.) If we could measure 
q, R, and H, we should at once have e/m. Now R can be meas- 



SEC. 13.4] MOTION OF A PARTICLE 387 

ured from a photograph of the track of the electron, and H can 
also be measured; but q presents a difficulty electrons move too 
fast for us to find their speeds directly. We have therefore to 
find q indirectly. Before entering the magnetic field, the 
electron is accelerated from rest by an electric field. If it starts 
from rest at potential V F and enters the magnetic field 
with speed q at potential V = Vi, then 



by the principle of energy (13.407). Elimination of q between 
the two equations gives 



This is a suitable expression for the determination of e/m, 
since all the quantities on the right are measurable. This 
is the method of Kaufmann. The electromagnetic units are 
such that (13.405) holds. 

Axially symmetric fields. 

Let R, <j>, z be cylindrical coordinates. A field is said to be 
axially symmetric with respect to the 2-axis if the potential is a 
function of R and z only (i.e., independent of <). An electric 
field of this type is produced by a system of charged plates 
perpendicular to the z-axis, a circular hole with center on the 
2-axis being cut from each plate. An axially symmetric mag- 
netic field is produced by currents flowing in circular coils 
arranged in planes perpendicular to the 2-axis, the centers of 
the coils being on the 2-axis. 

Let V be an axially symmetric electric potential. We assume 
that V can be expanded in a power series in x and ?/, the coeffi- 
cients being functions of z. On account of the axial symmetry, 
x and y can be involved only in the form x z + y 2 (= R 2 ), and so 
the expansion is of the form 

7?2 7?4 

(13.417) F = Fo(z) + g- Fi() + ~ 7,() + . 
Then, by an easy calculation, 

(13.418) a + 



388 MECHANICS IN SPACE [SEC. 13.4 

In order that Laplace's equation (13.402) may be satisfied, the 
functions F , FI, must satisfy the sequence of ordinary 
differential equations 

(13.419) FJ'OO + 27i(*) = 0, yj'W + |7 2 (s) = 0, . 

It is evident that VQ(Z) (the potential on the axis of symmetry) 
may be chosen arbitrarily, the burden of satisfying Laplace 's 
equations being placed on Vi(z), Vz(z\ . 

We may treat an axially symmetric magnetic field in exactly 
the same way. Assuming for the magnetic potential an expan- 
sion of the form 

P2 7?4 

(13.420) a = Goto + ~r Qi() + a 2 (*) + , 



we deduce the relations 

(13.421) OJ'CO + 2QiGs) = 0, ttJ'Cs) + iQ 2 (2) = 0, - - - . 

Motion of a charged particle near the axis of symmetry of an 
electromagnetic field. * 

If we introduce the potentials from (13.401) and (13.403), the 
general equations of motion (13.406) read, when written out in 
full, 



(13.422) 



e 



, , dV , . dfl . c.. 

a; = A- 1 -5- + y-; 2-5- 

x to * dz dy 

, ,'dV . . 50 . dti\ 

y = k (-^ + *te- x -te)' 



-k( a - 

~ k \dz 



*-^-y-^ 



where k = e/m. In the most interesting applications, the 
charged particle is an electron carrying a negative charge; in this 
case k is positive. 

Let us assume that the electromagnetic field has the 2-axis for 
axis of symmetry, so that we have the expansions (13.417) and 
(13.420) for V and 0, respectively. We shall consider only 
motion near the axis of symmetry, so that #, y, and their deriva- 
tives are small. Then, neglecting terms of order higher than the 

* Reference may be made to N. Chako and A. A. Blank, Supplementary 
Note No. I, in R. K. Luncberg, Mathematical Theory of Optics (Brown 
University, 1944) 



SEC. 13.4] MOTION OF A PARTICLE 389 

first, we rewrite (13.422) in the form 



y 

z = *FJ, 

where the prime denotes d/dz. The last of these equations is 
equivalent, in our approximation, to the equation of energy 
(13.407), which may be written 

(13.424) s 2 = 2k(V<> - C7), 

where C is a constant. This constant may be determined when 
the initial values of z and z are given. 

We note that (13.424) determines z as a function of z, to 
within a sign. Let us assume that z is positive throughout the 
motion. Then we may write 

(13.425) z = w(z) > 0, w* = 2k(V Q - C). 

The function w is the axial component of velocity. Since we are 
neglecting x z and ?/ 2 , it is clear that, to our order of approxima- 
tion, w also represents the magnitude of the velocity vector. 
By (13.419) and (13.421) we have 

(13.426) f V l = - 4*7 - - -L ^ (w*) - - ~ (urn/' + ^' 2 ), 

I Oi = - 40?. 

It is convenient to introduce complex notation, writing 
f = x + zy. We multiply the second equation of (13.423) by i 
and add it to the first; this gives, on making use of (13.426), 

(13.427) f = - \(ww" + u/ 2 )f - ftOjf - 4#ti>ni / r. 

We shall change the independent variable from t to z by the 
equations 

(13.428) f = fz = wf, f - u> s f" + tir. 
Substitution in (13.427) gives 

f r + JT + or = o, 

(13.429) _ ti/ 



"w" W "25"' 

This is the differential equation from which the path of the 
particle is to be determined by finding f as a function of z. 



390 MECHANICS IN SPACE [Sflc. 13.4 

The complex variable f represents the vector displacement of the 
particle perpendicular to the axis of symmetry (z-axis). The 
coefficients P and Q are functions of the independent variable z> 
and are determined by the axial potentials F (z) and tt (z) 
and by the initial conditions, which are needed to obtain the 
value of C in (13.425); we note that w depends on C. 

In the case of an electrostatic field, we put Q = 0; we note 
that then P and Q are real. In the case of a magnetostatic field, 
we put 7 = 0; then, by (13.425), w is a constant and the terms 
in P and Q involving derivatives of w disappear, leaving purely 
imaginary expressions. 

There is no simple general method of solving (13.429). How- 
ever, the equation may be simplified by using a standard device 
to eliminate the first-order derivative by a change in the depend- 
ent variable. To carry this out, we substitute 

(13.430) f(z) = u(z)v(z) 
in (13.429) and obtain 

(13.431) u"v + u'(2v' + Pv) + u(v" + Pv' + Qv) = 0. 

We now choose v so as to make the coefficient of u' vanish. We 
do this by writing 

(13.432) *> -expT-i f P) 

L J z o 

where z is the initial value of 2. When we substitute (13.432) 
in (13.431) we get, after some easy calculation, 



(13 433) 
(W.4A 

Also, by (13.430), 

(13.434) f = exp [ -* P(f) d*]. 

When we substitute for P and Q the expressions given in (13.429), 
we obtain a much simpler expression for S than we might expect, 
and (13.433) reads 

(13.435) " + 



, - 

The relation between the actual displacement vector f and the 



SBC. 13.4] MOTION OF A PARTICLE 

artificial displacement vector u is 

(13.436) f = 



391 



where W Q is the value of w when z = Z Q . 

It may be convenient for reference to write the results sepa- 
rately for the cases of electrostatic and magnetostatic fields: 

Electrostatic field: 

u" + S(z)u = 0, S(z) 



(13.437) 



W 



16 



w(z) = V2/f[Fo(z) - C], A- = - -, 

m 



Vo(Zo), WQ = W(Z ). 



Magnetostatic field: 

u" + S(z)u = 0, 

w o = constant velocity, 
(13.438) 



jj.20'2 

= it' 



6 

m 



f t/r 1 

= w exp ^- ( fi oo - flo) i 



There are some remarkable features in the preceding work. In 
(13.429) the coefficients P and Q were complex; but when we 
transform to (13.435), we get a real coefficient S. Not only is S 
real, it is also positive; this has an important bearing on the for- 
mation of " images" by an axially symmetric electromagnetic 
field, as we shall see later. 

The mathematical difference between the electrostatic case 
and the magnetostatic case is less than we might expect. In 
each case the coefficient S is positive. The chief difference lies 
in the relation between f and u. In the electrostatic case, the 
connection is real, and the complex vector f has the same direc- 
tion as the complex vector u. In the magnetostatic case the 



392 MECHANICS IN SPACE [SEC. 13.4 

connection is complex; the magnitudes of the two vectors are 
equal, and f is obtained from u by rotation through an angle 

5 <" - *> 

To sum up, in the general electromagnetic case, the determina- 
tion of the path of the particle involves the solution of (13.435). 
As initial conditions, we may assume that the particle starts 
from the point (20, f o) with velocity W Q in a direction giving to f ' 
the value fo- Thus, (13.435) is to be solved, with the initial 
conditions for z = ZQ, 

(13.439) u = f, ' - ft + iA- 
where 

no = n(*o), a = oj(*o). 

In the electrostatic case, (13.437) replaces (13.435), and we put 
floo = in (13.439); in the magnctostatic case, (13.438) replaces 
(13.435), and we put FH = in (13.439). 

Exercise. Show that the small angle between the initial velocity vector 
and the axis of symmetry is |fj|. 

The electromagnetic lens. 

In an optical instrument, such as a microscope, camera, or 
telescope, rays of light are bent by a system of glass lenses, 
the system usually having an axis of symmetry. The function 
of the instrument is to produce an image, the rays from each 
point of the object being brought to a focus at an image point. 
In recent times, there has been a remarkable development of 
electromagnetic devices analogous to the image-forming optical 
instrument. Instead of rays of light bent by glass lenses, there 
are streams of electrons whose trajectories are curved by means 
of electromagnetic fields. * When an axially symmetric electro- 
magnetic field is used, the equation (13.435) is the fundamental 
equation from which the trajectories of the electrons are 
determined. 

Let Oz (Fig. 139) be the axis of symmetry of an electromagnetic 
field; let n be a plane perpendicular to this axis, with the equa- 

* Cf. L. M. Myers, Electron Optics (Chapman & Hall, Ltd., London, 
1939), p. 100. 



SEC. 13.4] 



MOTION OF A PARTICLE 



393 



tion z = ZQ. Let P be any point on n with a complex position 
vector x + iy f . We suppose that from the point P Q there 
are projected a number of identical charged particles (electrons). 
Their velocities have a common magnitude w , but their direc- 
tions are different; the directions are, however, nearly parallel 
to the axis of symmetry, so that our methods apply. 

The trajectory of each electron satisfies (13.435). The func- 
tion w(z) is given by (13.425). Since all the electrons have the 
same charge e, the same mass m, and the same initial velocity w , 




FIG. 1JJ9 Formation of an image 

the constant C has the same value for all the trajectories. Hence 
w [and consequently 8 in (13.435)] is the same for all the trajec- 
tories. This is, of course, a mathematical idealization. As far 
as we know, all electrons have the same charge and the same mass, 
but we cannot secure accurately a common initial velocity. 
Hence, in practice, the constant C and the functions w and S 
will not be quite the same for all the trajectories. This leads 
to what is called, from the optical analogue, Achromatic aberra- 
tion," the velocity of the electron corresponding to the color of 
the light. But for our purposes we shall neglect this effect and 
regard w and S as the same for all the trajectories. 

We note that, from (13.439), for z = ZQ we have u = f for all 
the electrons; tmt the initial value of u f depends on the particular 
electron since f J is not the same for them all. 

It is known from the theory of linear differential equations 
that the general solution of (13.435) is of the form* 



(13.440) 



u = af(z) 



* Cf. E. L. Ince, Ordinary Differential Equations (Longmans, Green & 
Co., Ltd., London, 1927), p. 119. 



394 MECHANICS IN SPACE [SBC. 13.4 

where a, /? are arbitrary constants of integration (which may be 
complex) and /(z), g(z) are independent particular solutions. 
Since the coefficient S in (13.435) is real, we can obtain two real 
independent particular solutions by taking the initial conditions 

(13.441) /(z ) = 0, /'(z ) = 1; g(z ) = 1, g'(z Q ) = 0. 
With this choice of / and g, it follows from (13.440) that 

a = UQ, = u Q , 

where UQ, u' are the values of u, u' when z = z ; then (13.440) 
may be written 

(13.442) u = u'J(z) + u Q g(z). 

Let HI be the value of u at the point where the trajectory cuts the 
plane z = z\] then 



(13.443) u, = nJ/OsO + 

In the family of trajectories which we are considering, i.e., a 
family starting from a point (z , fo), u has a common value but 
UQ changes from trajectory to trajectory. Thus, in general, 
(13.443) will give an area on the plane z = z\ when we substitute 
the various values of u' Q corresponding to the various initial 
directions. The equation (13.443) will define a single point on 
the plane z = Zi if, and only if, 



(13.444) /fa) = 0. 

Let us recall that the function /(z) is defined by the following 
differential equation and initial conditions: 



(13 445) 

(13.445) /(* )=0, /(*,)-!. 

Can we find a plane z = z\ (other than z = z ) such that the whole 
family of trajectories cut it in a single point? This is equivalent 
to asking whether the equation (13.444) has a solution other than 

Zl = Z . 

Although we cannot give a definite answer to this question in 
general, we can discuss it qualitatively. Consider the graph of 
/(z). This graph starts from the z-axis at z , sloping up at 45. 
Thus/(z) is positive at first; hence by (13.445), since $ is positive, 
/"(z) is negative, and it remains negative as long as/(z) is positive. 



SEC. 13.4] 



MOTION OF A PARTICLE 



395 



This means that the graph is convex when viewed from above. 
Either of two things happens. The curve may turn down and 
cut the 2-axis at some point Zi (Fig. 140a); or it may turn so 
slowly that it reaches z = > before coming down to the 2-axis 
(Fig. 1406). In the former case, the equation (13.444) has a 
solution; in the latter case, it has no solution (at least not for 
values of z\ greater than z , and we are interested only in such 
values). In general terms, we may say that the larger S is, the 
more chance there is that there will be a solution; because the 
larger S is, the more rapidly does the graph of f(z) turn downward. 



f(z) 



f(z) 



o z 



(a) 



FIG. 140. Graph of f(z) : (a) in the case where an image is formcrl, (b) in the case 
wheie an image its not formed. 

If (13.444) has a solution, then the family of trajectories start- 
ing out from a point PO meet again in a point PI, as shown in 
Fig. 139. Borrowing the language of optics, we may call P the 
object point and PI the image point. In this sense, an axially 
symmetric electromagnetic field may form images. We might go 
further and say that it will form images if it is strong enough, 
because S is increased by an increase in the strength of the field. 

It will be noted that the equation (13.444), which determines 
the plane HI on which the image is formed, does not involve f . 
Consequently we may state the following important result: // 
an object point PO on the plane IIo has an image PI on the plane 
HI, then every object point on the plane n (near the axis of symmetry, 
to make the approximate method valid) has an image on the plane Hi. 
In fact, we have an object plane and an image plane, just as in the 



396 MECHANICS IN SPACE [SEC. 13.4 

optics of a lens. Hence, by analogy, we may speak of an axially 
symmetric electromagnetic field as an electromagnetic lens. 

Suppose that on the object plane n there is some minute struc- 
ture which we wish to photograph. We set up a photographic 
plate at the plane HI and bombard the plane n from the left 
with a stream of electrons. Each point of n becomes a source 
of electrons travelling on towards II i and converging to an image 
point on HI. Thus the structure on n is reproduced point for 
point on HI. A point on II transparent to electrons gives a 
" bright" point on Hi, and a point on HO opaque to electrons gives 
a "dark" point on IIj. Essentially, this is how an electron 
microscope works. Since magnification is the most important 
function of a microscope, let us now look into the question of the 
magnification m produced by an electromagnetic lens. 

The image of a point (z , fo) is (z\, f i), where Zi is given by 
(13.444) and fi by (13.436) and (13.443); we have 

(13.446) ui = u<*g(zi) = 

and 



(13.447) fl = f (*) exp 



[ - * ' 



where wi = w(zi). Magnification is defined by 

(13.448) m 

and so, by (13.447), the magnification of an electromagnetic 
lens is 

(13.449) m = 

We recall that g(z) is defined by the following differential equa- 
tion and initial conditions: 

(13.450) (jw-lf^^-O. 



It is a real function, since S is real; and the modulus sign in 
(13.449) is needed only to take care of the possibility that 
g(zi) is negative. 

In the case of an electrostatic field, the exponential disappears 
from (13.447); the vector fi has the same direction as f or the 



SEC. 13.4] MOTION OF A PARTICLE 397 

opposite direction, according as g(zi) is positive or negative. In 
the case of a magnetostatic field, (13.447) reads 

(13.451) ft = fo<7(zi) cxp ^~ (floo - Qoi)], 



where QOI is the value of Q at z = z\. The magnification is 
m = \g(zi)\. The image vector fi is obtained by first applying 
this magnification to the object vector f and then rotating it 
about the s-axis through an angle 

(13.452) = ^ (Goo - QOI). 

Figures 141a and b show the projections of object point PO, image 
point PI, and trajectories on the plane z = 0. They are drawn 
for m = 2 and a = 7r/4. In actual electron microscopes the 
magnification may be as high as 200,000. 

Approximations for electromagnetic lenses. 

The determination of the focal properties of an electromagnetic 
lens depends, as we have seen, on the solution of the differential 
equation 

(13.453) /"(z) + S(*)/CO = 0, 
with the initial conditions, as in (13.441), 

(13.454) /(z ) = 0, /'(so) = 1. 

There is no simple way of solving this equation, and we have to 
fall back on approximate methods. We shall consider the case 
where the electromagnetic field is concentrated on a short length 
of the z-axis, so that there is practically no field outside this short 
range. Making a mathematical idealization, we shall assume 
that there is a field for h < z < h and no field outside that 
range. 

In the absence of electric field, the axial potential V Q is con- 
stant, and so, by (13.425), w is constant and w' = 0. In the 
absence of magnetic field, the axial potential ft is constant and 
8 = 0. Thus, by (13.435), S = f or that range of values of z 
for which the electromagnetic field vanishes. By hypothesis, 
this is the case outside the range h < z < /?, and then the 
differential equation (13.453) becomes very simple: }"(z) 0. 



398 



MECHANICS IN SPACE 

y 



[SEC. 13.4 




FIQ. 14 la. Formation of an image by an electrostatic lens. 




Fio. 1415. Formation of an image by a magnetostatio lens. 



SEC. 13.4] 



MOTION OF A PARTICLE 



399 



The function f(z) is therefore a linear function of z. This 
corresponds to the fact that, in the absence of electromagnetic 
field, an electron travels in a straight line with constant velocity. 
The projections of trajectories in Figs. 14 la and 6 are drawn 
for such a case; each projection consists of two straight lines, 
connected by a curve. The curve is produced by the action 
of a concentrated electromagnetic field. If the field extended 



\45 



yy 

/A 
/,A 

I 

yy 

i 

^1 

// VV 

II 



S(z> 




z -h h z t 

Fia. 142. Graphs of S(z) and f(z) for a concentrated electromagnetic field. 

from z to 2i, the projections of the trajectories would be curved 
all the way. 

Figure 142 shows graphs of S(z) and /(z) for the case of a con- 
centrated electromagnetic field, drawn on the assumption that 
/(z) vanishes for some value Zi of z, so that an image is formed. 

Combining the initial conditions (13.454) with the fact that 
/(z) is linear for z < h, we obtain 

(13.455) /(-fc) - -h - Z Q , f(-h) = 1. 

Thus we know the value of / and its first derivative on entering 



400 MECHANICS IN SPACE [SEC. 13.4 

the concentrated field. We now try to find out what happens as 
we go through the field. 

Transferring the second term of (13.453) to the right-hand 
side and integrating from h to z, we get, remembering (13.455), 



(13.456) /'(z) = 1 - S(QJ(Q 
Another integration gives 

(13.457) /() = z - z - 



Putting z = h in these two equations, we obtain 

f /(/O = h - Z - f\ dr, f\ 8(&f(& dk, 
(13.458) { J J ~ h 



At first sight it may appear that we have found the values of / 
and its derivative on leaving the field, but of course this is 
illusory, because we do not know the function / occurring in the 
integrals. However, we can use the above equations as a basis 
for approximation. 

If S(z) is finite and h is small, it is evident from (13.458) that 
the changes in / and /' in passing from z htoz = h arc small. 
But to get an image, the graph of / must be bent through an 
angle of more than 45 on passing through the field. In fact, 
there must be a finite change in /', and consequently we must use 
a strong field. It is clear that S(z) must be large of the order 
h" 1 . Then the integral in the second of (13.458) is finite; the 
double integral in the first of (13.458) is small of order h, showing 
that although the change in /' is finite, the change in / is small. 

Before introducing the approximation, let us get an expression 
for Zi, the coordinate of the image point. From the linearity of 
the graph of / outside the field, we have 

(13.459) f(h) = - 

or 

^-- 

This will give us z\ if we can evaluate f(h) and/'(/0 from (13.458). 

We now take up the method of approximate solution by the 

method of iteration. The key equation is (13.457). For / under 



SEC. 13.4] MOTION OF A PARTICLE 401 

the sign of integration, we substitute / as given by the equation 
itself. This does not get rid of / on the right-hand side, but it 
pushes it under more signs of integration and thus reduces its 
importance. This procedure gives 



(13.461) /(*) -_.- dr, 8(& dt- * - 



dp]. 



The multiple integrals are to be evaluated starting from the right- 
hand side. We can rewrite this in the form 

(13.462) /() = 2 -0 + 2o f' h dr, J_' A 8(& d| 

h 



r dr, r s(& <* 

J h J h J 



This expression is accurate. For z in the range h<z<h, the 
first integral is small of order /i, and the remaining integrals are 
small of order h z . Differentiation of (13.462) gives 



(13.463) /'(z) = 1 + z S(Q dt- - z h &S(Q dl; 

f. S(Q d f f dq F S(p)f(p) dp. 

n J h J ti 



Here the first integral is finite, and the remaining integrals small, 
of order h. We observe that in (13.462) and (13.463), only the 
last integrals are unknown. 

If we required a higher approximation, we could substitute 
again for /under the sign of integration; then the integrals con- 
taining/would be small of order A 3 in the expression for /(z), and 
small of order h 2 in the expression for /'(z). This process could 
be continued indefinitely. 

Let us, however, content ourselves with approximations for 
f(h) and f(h) which retain terms of order h but reject terms of 
order ft 2 . We shall commit an error only of order /i 2 if we sub- 
stitute /(p) = z in the last integral in (13.463). Accordingly, 
putting z = h, we get the following approximate expressions: 



(13.464) 



f(h) =h-z + z h drj 
/'(/*) = 1 + z * S(S) d{ - _ $8(0 



dq S(p) dp. 



402 MECHANICS IN SPACE [SEC. 13.4 

To write these results more neatly, we introduce the finite 
constants 

f A = f S(Q dfc B = A-' / * 
(13.465) { J - h J ~ h 

[ D - *-' f $({) d f* (f - 

v %/ ~~ n %/ ~ n 

We note that, by inversion of order of integration, 



(13.466) 



Consequently, (13,464) read 

" z + /i[1 + (A ~ 



dq * S(p) dp 



The constants A, B, D may be evaluated numerically if the field 
is given; it should be noted however that they involve also the 
initial velocity MO, since S involves WQ (cf. equation (13.435)). 

Exercise 1. Show that if S = K/h, a constant, for h<z<h, then 

A - 2K, B = 0, D - ftf 2. 
Exercise 2. Evaluate 4, B, and Z), if, for -h <z <h, 

S(z) - a/i- 1 cos 2 1|, 
where a is a constant. 

Using (13.467) in (13.460), the image z\ corresponding to an 
object z is given by 



__ 

Zi _ h ~ ZQ ^ h[i + (A _ 

or more symmetrically, to the same order of approximation (i.e. 
neglecting /i 2 ), 

(13.469) --A = -h(- + - + 

Z\ ZQ \Zi ZQ 

If we let ZQ oo , we get 

(13.470) - A = -h(- + D\ 
z\ \z\ / 



SEC. 13.5] MOTION OF A PARTICLE 403 

The value of z\ so obtained gives the image of an object at 
infinity. 

If we are satisfied with the rougher approximation in which h 
is neglected, (13.469) becomes 

(13.471) - - - *, A. 

Z\ ZQ 

If we let Z Q > oo 9 the corresponding value of z\ is called the 
focal length F of the electromagnetic lens; by (13.471), we have 

(13.472) = A 

Exercise. Show that in the roughest approximation, the focal length 
of a concentrated electric lens is given by 



and the focal length of a concentrated magnetic lens by 



where H is the magnitude of the magnetic vector on the axis of symmetry 



13.6. EFFECTS OF THE EARTH'S ROTATION 

The effect of the earth's rotation on a plumb line was found 
in Sec. 5.3. This is a statical phenomenon relative to the rotating 
earth, and only the centrifugal force is involved. In dynamical 
problems on the rotating earth the Coriolis force also enters, and 
the effects are hard to predict without a careful mathematical 
analysis. 

Equations of motion of a particle relative to the earth's surface. 

We accept the model of the earth used in Sec. 5.3 an oblate 
spheroid turning about its axis of symmetry with constant 
angular velocity Q. The axis is supposed fixed in a Newtonian 
frame of reference. The vertical at any point on the earth's 
surface is defined by the plumb line, and the horizontal plane is 
perpendicular to the vertical. The latitude A is the angle of 
elevation of the earth's axis above the horizontal plane. 



404 



MECHANICS IN SPACE 



[SEC. 13,5 



In Fig. 143, SN is the earth's axis, drawn from south to north; 
is a point on or near the earth's surface, and B the foot of the 
perpendicular dropped from on SN] I is a unit vector along BO 
and K a unit vector parallel to SN. The triad of unit vectors 
i, j, k is fixed relative to the earth and directed as follows: 

i is horizontal and points south; 

j is horizontal and points east; 

k is vertical and points upward. 




]'io. 143. Vectors used in diHrussing the effects of the eaith'ts rotation. 

Let us put BO = a and denote by r the position vector of a 
moving particle relative to 0. Then the position vector of the 
particle relative to B is 

(13.501) r B = a + r. 

Since B is a fixed point in a Newtonian frame of reference, the 
absolute acceleration of the particle is 

(13.502) IB = + r. 

Here a is the acceleration of 0; since moves in a circle with 
constant angular velocity 12, we have 

(13.503) a == -aQ 2 I = -a!2 2 (sin X i + cos X k). 

Let m be the mass of the particle. The force of gravity is 
proportional to m and may be written mF. We denote by P 



SEC. 13.5] MOTION OF A PARTICLE 405 

the resultant of all other forces. The equation of motion is 

(13.504) mi B = wF + P, 
or, by (13.502) and (13.503), 

(13.505) mr = mF + P + mafl 2 (shi X i + cos X k). 

Let us apply this equation to a plumb line, hanging in equilib- 
rium with the bob at 0. Then P is the tension in the plumb line 
and points in the direction k. As in Sec. 5.3, we define g to be 
this tension, divided by the mass of the bob, so that 

P = mgk. 
Since r = 0, (13.505) gives 

(13.506) Fo + afl 2 (shi X i + cos X k) = -0k, 

where F is the force of gravity per unit mass at 0. 

Now F in the general equation (13.505) and F in (13.500) 
are not equal vectors unless the particle is at O, for the earth's 
gravitational field changes from point to point. But we shall 
assume that the particle always stays so close to that variations 
in the earth's field are negligible. So we introduce our first 
approximation, putting 

(13.507) F = Fo 

in the equation of motion (13.505). When, further, we substitute 
for F from (13.506), we get for any moving particle 

(13.508) mr = P - mgk. 

Our next task is to resolve this equation into components 
along i, j, k. This triad turns with a constant angular velocity 

(13.500) ft = OK = cos X i + ft sin X k, 

and so, by (12.310), 

(13.510) ? = S|+2QxJ + QX(OX f )' 

ot ot 

We now introduce a second approximation, dropping the last 
term on account of the smallness of 0. Substitution from 
(13.510) in (13.508) gives the vector form of the equations of motion 
relative to the earth's surface, 



406 MECHANICS IN SPACE [SEC. 13.5 

(13.511) m ~f = P - mgk - 2wQ X ~ 

Here 8*x/Bt* and dr/Bt are, respectively, the relative acceleration 
and velocity; the last term is the Coriolis force. The centrifugal 
force has been eliminated in two steps, first by (13.506) and 
secondly by neglect of the last term in (13.510). It will be 
noticed that (13.511) is essentially the same differential equation 
as (13.406) or (13.408) the equation of motion of a charged 
particle in an electromagnetic field. 

Let us now introduce axes Oxyz coincident in direction with 
(i, j, k), so that Ox points south and Oy east. Let X, F, Z be 
the components along these axes of the force P, which, it will 
be remembered, is the force other than gravity. Since Q is 
given by (13.509), we get from (13.511) the scalar form of the 
equations of motion, 

Imx = X + 2wQ sin X y, 
my = Y - 2mO(sin X x + cos X z), 
mz = Z mg + 2m$l cos X y. 



Motion of a free particle. 

By a "free particle" we mean here a particle on which there 
acts no force but gravity. As remarked in Sec. 6.1, this is an 
idealization difficult to approach in practice. The resistance 
of the air is always present and produces discrepancies between 
mathematical predictions and observed motions. It is therefore 
not surprising that the minute effects due to the earth's rotation 
are hard to detect. 

For a free particle, we put X = Y = Z = in (13.512). The 
resulting equations are easy to integrate, especially as further 
approximations, based on the smallness of 8, are permissible. 
The equations now read 



(x 2ft sin X y, 
y = -2Q(sin X : 
z = -.0 + 28 GOJ 



(13.513) \ y = -2Q(sin X + cos X z), 

t cos X y. 

Each of these equations can be integrated once. Without loss 
of generality, we may suppose that the particle starts from the 
origin at t = with velocity (u , t>o, Wo), and so we get 



SEC. 13.5] MOTION OF A PARTICLE 407 

Ix ~ 2J2 sin X y + UQ, 
y = -2Q(sin X x + cos X z) + t> , 
z = gt + 212 cos \ y + WQ. 

If we substitute from the first and third of these equations in 
the second of (13.513) and neglect 12 2 , we obtain 

(13.515) y = 212(^0 sin X + WQ cos X gt cos X), 
and hence, by integration, 

(13.516) y = v t lM 2 (w sin X + w cos X) + i%* 3 cos X. 
Then the first and third equations of (13.514) give, on neglecting 



x ** Uot + Vot * sin x> 

z = wot- \gt^ + Qv Q t* cos X. 



Two cases are of particular interest, a particle dropped from 
rest, and a particle representing a projectile fired with high 
velocity in a flat trajectory. 

In the case of a particle dropped from rest, we put 

UQ = VQ = Wo 0, 

and get 

(13.518) x = 0, y = $Qgt* cos X, z - - $gt*. 



The path is a semicubical parabola in the east-west vertical 
plane, 

(13 . 5 19) , Z= -.i. z . 



It is evident from (13.518) that the deviation from the vertical is 
toward the east. From (13.519), the deviation for fall from a 
height h is 



fl cos X - 2h . 
\ y 

This is zero at the poles (X = ^r), as we should expect. 

For a projectile with large UQ and Vo, we neglect the term in 
WQ and also the last term in (13.516). Thus the projection of the 
trajectory on the horizontal plane has the equations 

(13.520) x = Uot + Itoo* 2 sin X, y = v Q t - Quo? sin X. 



408 MECHANICS IN SPACE [SEC. 13.5 

These may be expressed in complex form in the single equation 
(13.521) x + iy = (U Q + iv Q )(t - ilM* sin X). 

Let us put 

x + iy = Re*+, UQ + iv = qoe ia 
and write, as we may since ft is small, 

1 Hit sin X = exp (Hit sin X). * 
Then (13.521) takes the form 

q l exp (ia iQ,t sin X), 



and so 

(13.522) R = got, <t> = a ft sin X. 

The magnitude of the position vector grows at a constant rate go, 
and at the same time the vector turns at a constant rate sin X. 
In the Northern Hemisphere, X is positive and this rotation is 
clockwise when viewed from above; in the Southern Hemisphere, 
it is counterclockwise. This means that the projectile experi- 
ences, on account of the earth's rotation, a slight deviation 
to the right in the Northern Hemisphere, and to the left in the 
Southern Hemisphere. This is known as FereVs law. 

Foucault's pendulum. 

Let us suppose a pendulum set up at the North Pole. If 
started properly, it may vibrate as a simple pendulum in a 
vertical plane which is fixed in the Newtonian frame of reference. 
As the earth turns under the pendulum with angular velocity 12, 
the plane of vibration of the pendulum appears to an observer 
on the earth to turn with an angular velocity ft. Foucault 
was the first to point out that a pendulum could be used to 
demonstrate the earth's rotation. It is not necessary that the 
pendulum should be situated at one of the earth's poles; an 
apparent rotation, due to the rotation of the earth, may be 
observed at any latitude except on the equator. 

We shall now apply (13.512) to the motion of a pendulum. 
The pendulum consists of a particle of mass m attached by a 
light string of length a to a point with coordinates (0, 0, a). 
Thus, in equilibrium the particle rests at the origin. We shall 
discuss small oscillations about this position, a problem already 



SEC. 13.5J MOTION OF A PARTICLE 409 

solved [cf. (13.304)] for the case 12 = 0. The question of interest 
now is to find how the simple motion there described is modified 
by the rotation of the earth. 

We recall that X, Y, Z are the components of force other 
than gravity. For the pendulum, this consists of the tension 8 
in the string; as in (13.302) the components are 

(13.523) Z = - - S, 7 - ^ S, z = 5LZ^ s. 

CL a QI 

We have to take care of two separate approximations. The 
first, based on the smallness of 12, has already been used in 
obtaining (13.512); it consists in neglecting 12 2 . The second 
approximation is that arising from the smallness of the oscilla- 
tions. This means that x, y, and their derivatives are small; 
z and its derivatives are therefore small of the second order and 
consequently will be neglected. 

The last equation of (13.512) gives, since Z = S approxi- 
mately, 

(13.524) S = mg 2ml2 cos X y, 
and so the first two equations become 

~ 20 sin X ' ^ + p ' x ~ > 
+ 212 sin X z + p*y = 0, 

where p* g/a. Multiplying the second equation by i and 
adding it to the first, we get the single complex equation 

(13.526) f + 2tl2 sin X f + p^ = 0, (f = x + iy). 
The general solution is 

(13.527) f = A&*< + Be n , 

where A and B are complex constants depending on the initial 
conditions and HI, n^ are the roots of the equation 

(13.528) n 2 + 2iQ sin X n + p* = 0. 
These roots are 



ni, n 2 = t'O sin X i \/ti 2 sin 2 X + 
Neglecting 12 2 , we may write (13.527) in the form 
(13.529) f = f i exp (-i$lt sin X), 



410 MECHANICS IN SPACE [SEC. 13.5 

where 

(13.530) f i = A# + Be-*". 

To interpret this result, we suppose for the moment that 12 = 0, 
so that f = f i. On separation of the real and imaginary parts, 
it is easily seen that the path is an ellipse with center at the 
origin the motion being a composition of perpendicular simple 
harmonic motions [cf. (6.405) and (13.304)]. The effect of the 
second factor in (13.529) is to rotate the complex vector fi 
through an angle tit sin X, proportional to the time. We may 
sum up our result as follows: The effect of the earth's rotation 
on the elliptical path of a spherical pendulum is to cause the ellipse 
to rotate with an angular velocity ft sin X. This rotation is 
clockwise in the Northern Hemisphere and counterclockwise in 
the Southern Hemisphere. If we put X = far, so that the pen- 
dulum is at the North Pole, the angular velocity becomes ft 
and may be regarded as due directly to the earth's rotation 
beneath the pendulum. 

When we discussed the spherical pendulum in Sec. 13.3, with- 
out taking the earth's rotation into consideration, we started 
with a first approximation and obtained an elliptical orbit from 
equations (13.304). We then proceeded to a second approxi- 
mation and found, in (13.333), an expression for the rate at which 
the elliptical orbit advances. In the case of Foucault's pendu- 
lum, the situation is more involved. In the first place, the 
angular velocity ft of the earth is small (cf. page 143), and that 
fact was used in obtaining (13.512). Secondly, we have con- 
sidered only small oscillations (first approximation). If we were 
to proceed to a second approximation, we would find two super- 
imposed rotations of the elliptical orbit one depending, as in 
(13.333), on the area of the orbit (area effect), and the other an 
angular velocity 12 sin X due to the earth's rotation (Foucault 
effect). Unless special precautions are taken, the area effect is 
likely to be much larger than the Foucault effect and to conceal 
it. To prevent this, it is usual to draw the pendulum aside with 
a thread and start the motion by burning the thread. This 
means that, for t = 0, we have f = and f = f o (say). Then, 
by (13.529), 

(13.531) A + B = fo, p(A - JB) - (A + 7?) ft sin X - 0. 



SBC. 13.6) MOTION OF A PARTICLE 411 

Now (13.530) may be written 

(13.532) f i = (A + B) cos pt + i(A - B) sin pt, 
or, by (13.531), 

(13.533) fi = ft (cos pt + i sin pt - sin X/p). 

It is easy to see that this represents motion in an ellipse with 
semiaxes |fo| and \fo\Q sin X/p. For an ellipse with these semi- 
axes, described in a Newtonian frame of reference, the advance 
of the apse in one period (2ir/p) is, by (13.333), 

(13.534) ~^- |f ol'ft sin X, 

and so the angular velocity of the ellipse due to the area effect is 

(13.535) ^ sin X. 

Since |f o|/a is small, this angular velocity is much smaller than 
the Foucault angular velocity ft sin X and may be regarded as 
negligible. Hence, if the pendulum is started by burning a 
thread, the rotation of the orbit may be regarded as due to the 
Foucault effect alone. 

It should be noted that, for any small elliptical orbit, there 
is a distinction between the area effect and the Foucault effect. 
The area effect is always a rotation in the sense in which the 
ellipse is described and reverses when that sense is reversed, but 
the Foucault rotation takes place in a definite sense (clockwise in 
the Northern Hemisphere and counterclockwise in the Southern). 

13.6. SUMMARY OF APPLICATIONS IN DYNAMICS 
IN SPACE MOTION OF A PARTICLE 

I. Jacobian elliptic functions. 
(a) Differential equation: 



(13.601) gay = (i - ifld - *V), (0 < * < i). 

(b) General solution: 

(13.602) y = sn (a: + c). 



412 MECHANICS IN SPACE [SEC. 13.6 

(c) Other elliptic functions: 

(13.603) en 2 x = 1 - sn 2 x, dn 2 x = 1 - P sn 2 x\ 
(13.004) -7- sn x = en x dn x, j- en x = sn x dn z, 

-r- dn # = A 12 sn a: en a;. 
dx 

(d) Periodicity: 

(13.605) sn (x + 4/0 = sn re, en (x + 4/0 = en a;, 

dn (x + 2/0 = dn x; 

jc r_^ = JL===,= p- , d * .. 

JO ^/(i _ y 2)(! _ ^.2^2) JO ^ J2 gin 2 ^ 



II. Simple pendulum. 

(a) Motion: 

I sin ?6 sin ^a sn [p(t to)], 
, g . - i 
^2 , /,. = gm ^ a 

(b) Periodic time: 
(13.608) r- 



- A- 2 sin 2 
- (1 + iVa 2 ), approximately. 

III. Spherical pendulum. 

(a) General motion: the pendulum oscillates between two 
levels, found by solving a cubic equation; the analytical solution 
is 

It - zi = (2 2 - Zi) sn 2 [p(t - <<,)], 
= ?, F = g2 ~ g ^ 
a 3 - 21 

(6) Small oscillations: in the first approximation the path is 
an ellipse; in the second approximation the ellipse turns at a 
rate proportional to its area. 

IV. Motion of a charged particle in an electromagnetic field. 

(a) Uniform electric field: the trajectory is a parabola. 
(6) Uniform magnetic field: the trajectory is a helix. 



HEC. 13.6] 



MOTION OF A PARTICLE 



413 



(c) Axially symmetric electromagnetic field: the trajectory 
satisfies 



u" + S(z)u = 0, 



3F' 2 



16 



(13.610) 



- * + % = exp - 



> 
iw. 



= speed of particle = ^/2Ar f F F o -f r 



Fo = axial electric potential, F o = Vo(z ), 
fio = axial magnetic potential, 
Prime indicates d/dz. 



(d) Electromagnetic lens: 

(i) Image plane z z\ of object plane z = 2 determined by 



(13.611) 



= 0, 
= 0, 

(ii) Magnification given by 



= 0, 



(13.612) 



o, 



o, 



w = 



(e) Magnet ostatic lens : rotation of image given by 



(13.613) 



414 MECHANICS IN SPACE [Ex. XIII 

V. Effects of the earth's rotation. 
(a) Equations of motion: 



(13.614) 



mx = X + 2mQ sin X y y 

my = Y 2wl2(sin X x + cos X z), 



mz = Z mg + 2mfl cos X y, 

X, Y, Z = force other than gravity, 

X = latitude. 



(6) A falling body deviates to the east. 

(c) A projectile deviates to the right in the Northern Hemis- 
phere. 

(d) Foucault's pendulum turns with angular velocity 12 sin X, 
clockwise in the Northern Hemisphere. 

EXERCISES XIII 

1. A simple pendulum of mass m and length a performs finite oscillations, 
the greatest inclination of the string to the vertical being 30. Find the 
tension in the string when the bob is in its highest position. 

2. The string of a spherical pendulum is held out horizontally and the 
bob started with a horizontal velocity perpendicular to the string. Find to 
the nearest foot per second the magnitude of this velocity if the minimum 
inclination of the string to the vertical in the subsequent motion is 45. 
The length of the string is 54 inches. 

3. A particle carrying a charge e is projected from the origin with a 
velocity UQ in the direction of the z-axis. There is a uniform magnetic field 
of strength H parallel to the z-axis. If the particle crosses the plane x = 
at a distance a from the origin, find its mass. (Isotopes are separated in a 
mass spectroscope by a method such as this.) 

4. A heavy bead is free to move on a smooth circular wire of radius a, 
which rotates with constant angular velocity Q about a fixed vertical diam- 
eter. Find the possible positions of relative equilibrium. If 12 2 > g/a t 
find the period of small oscillations about a position of stable equilibrium. 

5. A stone is thrown straight up, rises to a height of 100 ft., and falls to 
earth. Estimate the deviation due to the rotation of earth, the latitude of 
the place being 45 North. (Neglect air resistance.) 

6. A particle moves under gravity on a smooth surface of revolution 
with axis vertical. The equation of the surface in cylindrical coordinates is 
given in the form R =* F(z). If the velocity is horizontal and of magnitude 
q\ at a height z\, and again horizontal and of magnitude q<i at a height Zi, 
determine qi and # 2 in terms of zi and ^^. (Use the principles of energy and 
angular momentum.) 

7. In a simple pendulum the bob is connected by a light string of length a 
to the fixed point of support. The bob starts in the lowest position with 
speed g . Show that if 



Ex. XIII] MOTION OF A PARTICLE 415 

the string will slacken during the motion, so that the bob falls inward from 
the circular path. 

8. Show that, on account of the rotation of the earth, a train traveling 
south exerts a slight sideways force on the western rail of the track. Give 
an approximate expression for this force in terms of the mass of the train, its 
speed, the latitude, and the angular velocity of the earth. 

9. A particle moves on a smooth surface of revolution with axis vertical. 
The equation of the surface in cylindrical coordinates is R F(z). Use the 
principles of angular momentum and energy to show that 

R*4> = h, 
%(z* + & + RW) -h gz - E, 

where h and E are constants. Deduce that z satisfies a differential equation 
of the form 

& <=/(*). 

10. A particle slides on a smooth cycloid in a vertical plane, the cusps of 
the cycloid being upward. Show that the periodic time of oscillations under 
gravity is independent of the amplitude. 

11. A spherical pendulum of length a and mass ra oscillates between two 
levels which are at heights b and c above the lowest point of the sphere. 
Express its constant total energy in terms of m t a, 6, c, and g, taking the zero 
of potential energy at the lowest point of the sphere. Check your answer by 
putting b = c. 

12. A charged particle moves in a uniform electric and magnetic field, 
the electric and magnetic vectors being perpendicular to one another. 
Show that, if properly projected, the path of the particle is a cycloid. 

13. The bob of a spherical pendulum, 10 feet long, just clears the ground. 
A peg is set up 1 foot due south from the equilibrium position of the bob, and 
the bob is drawn out to the east through a distance of 2 feet. Find (approxi- 
mately) the direction and magnitude of the velocity with which the bob 
should be started from this position, in order to hit the peg and make it fall 
over toward the west. 

14. For a simple pendulum of length a making complete revolutions, show 
that the periodic time is 



4a fl 

: I 

qo JO 



dy 



- 2/ 2 )U - 



where q Q is the speed at the lowest position and k* =* ql/4ga. 

16. A smooth cup is formed by revolution of the parabola z a 4ax about 
the axis of z, which is vertical. A particle is projected horizontally on the 
inner surface at a height z with a speed -\/2kgz Q . Prove that, if A; = i, the 
particle will describe a horizontal circle; also that, if k F J 0j its path will lie 
between two planes z = z and z = |z . 

16. A spherical pendulum of length a is held out horizontally, and the bob 
is started with a great horizontal velocity q Q . Show that it falls to a depth 
below its initial position given approximately by 2gra 2 /c5. 



416 MECHANICS IN SPACE [Ex. XIII 

17. A particle moves on smooth surface of revolution, the axis of symmetry 
being vertical. Show that motion in a horizontal circle of radius R is stable 
if 

d*z 3 dz 



where z = z(R) is the equation of the surface in cylindrical coordinates. 
Deduce that the motion of a conical pendulum is stable. 

18. A particle moves under gravity on a rough vertical 'circle. It starts 
from rest at one end of the horizontal diameter and comes to rest at the 
lowest point of the circle. Find an equation to determine the coefficient of 
friction. 

19. A heavy bead starts from rest at a point A and slides down a smooth 
wire in. the form of a helix having the parametric equations 

x a cos 0, y = a sin 0, z = aO tan a, 

the axis of z being vertical. When it is vertically under A, a second bead 
starts from rest at A. Show that the tangential acceleration of each bead is 
g sin a, and deduce a formula determining all the subsequent instants at 
which one bead is vertically underneath the other. 

20. A heavy particle is constrained to move on the inner surface of a 
smooth right circular cone of semivcrtical angle a, the axis of the cone being 
vertical and the vertex down. The particle is in steady motion in a circle 
at height b above the vertex. Find the periodic time for small oscillations 
about this steady motion. 

21. A particle slides in a smooth straight tube which rotates with con- 
stant angular velocity about a vertical axis which does not intersect the 
tube. The tube is inclined at an anglo a to the vertical. Initially the par- 
ticle is projected upward along the tube with speed <?o, relative to the tube, 
from the point where the shortest distance between the axis and the tube 
meets the tube. Show that, no matter what the length of the tube may be, 
the particle will escape at the upper end provided 



co tan a 

22. Consider an axially symmetric electric field in which the axial poten- 
tial is of the form Fo = az + & Show that the field is uniform through- 
out space and parallel to the 2-axis. Verify directly from (13.429) that the 
trajectory of a charged particle is parabolic. 

Consider also the case of an axially symmetric magnetic field with the 
axial potential of the form Q = az + b. Verify from (13.438) that the 
trajectory is a helix. 

23. A particle moves under gravity on a smooth surface, the principal 
radii of curvature at its lowest point being a, b (a > ft). The surface rotates 
with constant angular velocity w about the normal at the lowest point. 
Show that, if Oxy are horizontal axes attached to the surface at the lowest 
point and directed along the lines of curvature at that point, then the equa- 



Ex. XIII] MOTION OF A PARTICLE 417 

tion of motion for small vibrations near the lowest point are 



Considering solutions of the foim x = Ae nt , y - Be nt , deduce that there will 
be instability if w* lies between g/a and g/b. 

24. If in a magnetostatic lens the axial component ft of the magnetic 
vector is constant, show that an object point on tho axis at z = will give 
an image at 



z = __, 

where WQ is the initial velocity. 

25. For a magnetostatic lens, with the field concentrated in h < z < h t 
prove that, to the order h inclusive, the magnification of an object in the 
plane z Z Q is 

|l +ylzo - k(B + Dzo)h, 

where A, #, D aie constants defined by (13405) Evaluate explicitly, if 
the axial component of magnetic field H is constant in h < z < h. 



CHAPTER XIV 



APPLICATIONS IN DYNAMICS IN SPACEMOTION OF A 
RIGID BODY 

14.1. MOTION OF A RIGID BODY WITH A FIXED POINT UNDER 
NO FORCES 

If a rigid body is constrained to turn about a smooth fixed 
axis, under no forces other than the reaction of the axis, the 
motion is extremely simple: the body spins with constant angular 
velocity. But if, instead of fixing a line in the body, we fix 
one point only, the motion under no forces is much more com- 
plicated. The problem of finding this motion is of wider interest 

than might appear at first sight, for 
the motion of a rigid body relative 
to its mass center is the same as if 
the mass center were fixed (cf. Sec. 
12.4). 

The mounting of a body so as to 
fix only one point is much more 
complicated than that required to 
give it a fixed line. It may be done 
by an arrangement of light rings, 
known as "Cardan's suspension " 
(Fig. 144). The body is represented 
by the inner circle. The points A, B 
are fixed. Rotation of the ring R i 




FIG. 144. Cardan's suspension. 



about A B gives one degree of freedom. Rotation of the 
ring R 2 about CD gives a second degree of freedom. Rotation 
of the body itself about EF gives the third. The body can take 
up all positions in which the point of the body is fixed in space, 
being the common intersection of AB, CD, and EF. All the 
apparatus, except the body itself, is to be regarded as massless 
in the mathematical theory; this cannot, of course, be achieved 
in practice, but the masses of Ri and R% are made as small as 
possible compared with the mass of the body. 

418 



SEC. 14.1] MOTION OF A RIGID BODY 419 

There are two ways of treating the problem of the motion of 
a body with a fixed point under no forces the descriptive and 
the analytic. The descriptive method, or method of Poinsot, 
gives a good qualitative idea of the motion. In the case where 
the body has an axis of dynamical symmetry, the description is 
particularly simple; we shall consider that case in detail later. 

The method of Poinsot. 

Let be the fixed point in the body, and A t B, C the principal 
moments of inertia at 0. Let i, j, k be unit vectors fixed in the 
body and directed along the principal axes at 0. For the angular 
velocity and angular momentum we have, by (11.509), 



(14.101) to = cjii + co 2 j + w 3 k, h = 

When we say that the body is under no forces, we mean more 
precisely that the forces acting on 
the body have no moment about 0. 
(Thus our argument applies to a 
heavy body under the action cf 
gravity, provided that is the 
center of gravity.) Since the ex- 
ternal forces do no work and have 
no moment about 0, we have the 
following facts to assist us in dis- 
cussing the motion : 

/\ ji_ i j. m FIG. 145. Tho invariable line 

(l) the kinetic energy T IS and the invariable plane. 

constant; 

(ii) the angular momentum h is a constant vector. 
From the first of these we have, by (11.404), 




(14.102) Aco? + Bu\ + Cul = 2T = constant; 



from the second, we know that h has a direction fixed in space 
and also a constant magnitude, so that 



(14.103) A 2 co? + 2 co| + C 2 l = h* = constant. 

Let us draw through a line OP in the fixed direction of h 
(Fig. 145); this is called the invariable line. Let OQ represent 
the angular velocity <> at any instant. Drop the perpendicular 
QN on OP; then ON = o> h/h. But, by (14.101) and (14.102), 

(14.104) <* h - 2T, 



420 MECHANICS IN SPACE [SEC. 14.1 

and so 

2T 

(14.105) ON = ~ = constant. 

Thus N is a fixed point during the motion, and so the plane 
through N, perpendicular to the invariable line OP, is a fixed 
plane; it is called the invariable plane. The extremity Q of the 
angular velocity vector co moves on the invariable^plane. 

Let us now take the point of view of an observer who moves 
with the body. (This is what we do in our daily lives, for we live 
on a rotating earth but regard a point on the earth's surface as 
" fixed.") To such an observer, the vectors i, j, k are fixed, but 
both the vectors h and CD are changing. If i, j, k are taken as 
coordinate axes and the extremity of the vector co is given 
coordinates x, y, z, then 

x = on, y = o) 2 , z = o) 3 . 
By virtue of (14.102) and (14.103), we have 

(14.106) Ax 2 + By 2 + Cz 2 = 2T, A*x* + B 2 y 2 + C 2 z* = h 2 . 

In fact, to an observer moving with the body, the extremity Q of 
the angular velocity vector o describes a curve which is the inter- 
section of the two ellipsoids (14.106), fixed in the body. 

The first of these two ellipsoids is similar to the rhomental 
ellipsoid and has the same axes; it is called the Poinsot ellipsoid. 

The invariable plane is fixed in space, but to the observer 
moving with the body it is a moving plane. It touches a sphere 
of radius ON, but it has another remarkable property: the 
invariable plane touches the Poinsot ellipsoid at the extremity of 
the angular velocity vector. 

To see this, we note that the tangent plane to the Poinsot 
ellipsoid at the point (on, o> 2 , o> 3 ) is 

(14.107) AU& + B^y + CW = 2T. 



So the direction ratios of the normal to the ellipsoid at this 
point are 

A&I, Buz, C3. 

But these are precisely the components of angular momentum; 
hence the normal to the Poinsot ellipsoid at the extremity of the 
angular velocity vector is parallel to the angular momentum 
vector, i.e., parallel to OP. This proves the result. 



SEC. 14.1) MOTION OF A RIGID BODY 421 

As we have indicated, there are two different points of view: 

(i) the point of view of an observer S fixed in space; 

(ii) the point of view of an observer S r fixed in the body. 
It is confusing to try to look at things simultaneously from the 
two points of view. We shall clarify the situation by taking 
them up separately. 

The observer /S, fixed in space, cuts away (in his imagination) 
all the body except an ellipsoid the Poinsot ellipsoid. He 
fixes his attention on this moving ellipsoid and on a fixed plane 
(the invariable plane). As the body moves, the ellipsoid always 
touches the plane. It actually rolls on the plane, since it has 
an angular velocity vector which passes through the point of 
contact of the ellipsoid and the plane. This is a fairly com- 
plicated type of motion; it becomes quite simple, however, 
when the Poinsot ellipsoid is a surface of revolution, as we shall 
see later. But it may in any case be visualized by thinking of 
the invariable plane as a sheet of paper and the Poinsot ellipsoid 
as an inked surface. In the course of the motion a curve is thus 
drawn in ink on the invariable plane. On joining the fixed point 
to the points on this curve, we get the space cone (cf. Sec. 11.2). 

On the other hand, the observer /S', fixed in the body, turns 
his attention to the two ellipsoids (14.106), fixed as far as he is 
concerned, and in particular to their curve of intersection. The 
angular velocity vector traces out a cone (the body cone), formed 
by joining the fixed point to this curve. 

The two points of view are brought into contact by the general 
result: the body cone rolls on the space cone. The difference 
between the two is this: S regards the space cone as fixed, but 
S' regards the body cone as fixed. 

The above method gives a qualitative, rather than a quantita- 
tive, description of the motion. For a quantitative description, 
we must use an analytic method. 

The case of a general body; analytic method. 

Since the external forces have no moment about 0, Euler's 
equations (12.404) give 

!Ai - (B - C)w 2 o> 3 = 0, 
Ba 2 - (C - A)co 3 o>i = 0, 
Cw 3 - (A - #)!, = 0. 



422 MECHANICS IN SPACE [SBC. 14.1 

We have also, as in (14.102) and (14.103), 

(14109) 
U4.iuy; 



where T and h are constants, which may be found by inserting 
the values of i, w 2 , w 3 at = 0. [The equations (14.109) may 
be deduced from (14.108) directly.] 

We shall assume that A, B, C are all distinct. We may suppose 
the triad i, j, k chosen so that A > B > C. Then it follows 
from (14.109) that 

2AT - h* > 0, 2CT - h* < 0. 

There are three very simple particular solutions of (14.108). 
These are 

o>i = constant, W2 = 0, s = 0, 
to 2 = constant, w 3 = 0, coi = 0, 
wa = constant, wi = 0, W2 = 0. 

These three solutions correspond to steady rotations about the 
three principal axes of inertia. It is a remarkable fact that these 
are the only axes about which the body will spin steadily 
under no forces; the equations (14.108) are satisfied by constant 
values of i, o>2, o>a only if two of these constant values are zero. 
Turning now to the problem of finding the most general solu- 
tion of (14.108), we must first eliminate two of the unknowns, 
so as to get a differential equation involving just one unknown. 
It proves best to concentrate our attention on w 2 . We solve 

(14.109) for wf, J, obtaining 

(14.110) o>? = P - Qcol, wS = R - SJ, 

where P, Q, R, S are positive expressions involving A, B, C, T, h. 
Substitution in the second equation of (14.108) gives 



This equation is of the same form as (13.312) and may be treated 
in the same way. Thus, from (14.111), we get 



where 
(14.113) 



SEC. 14.1] 



MOTION OF A RIGID BODY 



423 



the constants 0, p, k being positive functions of A, B, C, T, h, with 
k < I. Hence, 



(14.114) 



= sn [p(t - to)], 



where t Q is a constant of integration. Substitution in (14.110) 
gives either 



(14.115a) 

or 

(14.1156) 



a dn [p(t - <)], 



a en [p( - 



a> 3 



7 en [p(t - )], 



7 dn [p(l - )], 



where a and 7 are functions of A, B, C, T, h, determined except 
for sign. When we substitute in (14.108), we find that ajfry is 
negative. For definiteness, we may make a positive by suitable 
choice of the sense of the vector i; then 7 is negative. 

That is the outline of the method of finding coi, o> 2 , o> 3 as func- 
tions of t. The completion of the argument consists in filling 
in the algebraic details. Care must be taken in selecting the 
constants /?, p, k, so that k is less than unity; it becomes necessary 
to distinguish between the two cases (a) ft 2 > 2BT, and (6) 
h* < 2BT. In the former case, we arrive at (14.115a), in the 
latter at (14.1156). We leave it to the reader to verify the 
following results. 

CASE (a): h' 2 > 2BT. 



CASE (6): 
(14.1166) 


//i 2 - 2CT 


1 




a \A(A - cy 


(A 2 - 2CT)(A - B) 


12 A T - /i 2 


/ 


ABC 


^ ^B(A - By 


B - C 2 AT - h 2 


<\t 1 i 


^ ^ 

p=V 

-V 


A - B h 2 - 2CT 


7 " \lc(A - cy 

h 2 < 2BT. 


lh* - 2CT 


a ^A(A - C)' 


(2A7 1 - h 2 ) (B - C) 


l h * - 2CT 


AJ3C 


P ~ *\B(B - C)' 


A - B ft 2 - 2CT 


I2AT - h 2 

7 m~m I ----,,, . 


B - C 2AT - ^ 


\\r( A rV 

\ i/^/i uy 



424 MECHANICS IN SPACE [SEC. 14.1 

In establishing these results, the following identity is useful: 



(14.117) (B - C)(^ 2 - 2AT) + (C - A)(h* - 2BT) 

+ (A - B) (h* - 2CT) = 0. 

But the determination of i, co 2 , w 3 as functions of t does not 
complete the solution of the problem. We should be able to 
tell, from given initial conditions, the position of the body at any 
time. To do this, we specify the directions of the triad i, j, k, 
relative to a triad I, J, K, fixed in space, by means of the Eulerian 
angles 0, *, ^. Then, by (11.202), 

!o>i = sin \l/ 6 sin 9 cos ^ <, 
co 2 = cos $ 6 + sh) sm ^ <, 
0) 3 = COS + $. 

If we substitute for OH, co 2 , u 3 from (14.114) and (14.115), we 
obtain three differential equations for 6, <, ^. The solution of 
these equations in this general form presents a formidable 
problem. It is greatly simplified if we choose the vector K in the 
direction of the invariable line, defined by the constant vector h. 
Then the components of h along i, j, k are found by multiplying 
h by the direction cosines of K relative to i, j, k. Those direction 
cosines are easily found (cf. Fig. 118, page 280) by projecting K 
on i, j, k; they are 

sin cos ^, sin sin ^, cos 6. 
Hence, 

!Ao>i = h sin 6 cos ^, 
Bw z = h sin 6 sin f , 
Co> 3 = h cos 6. 

From these equations, we get 6 and \l/ as functions of t without 
any integration, thus: 

(14.120) cos e = ^p, tan * = - |^- 

To find <f>, we deduce, from the first two of (14.118), 

(14.121) sin (j> = co 2 sin \f/ o>i cos ^, 

and so <f> is obtained by a quadrature, since 0, ^, wi, co 2 are already 
known as functions of t. 



SEC. 14.1] MOTION OF A RIGID BODY 425 

From the periodic property of the elliptic functions, we see 
that 0, sin ^, cos ^, < are periodic functions of /; in general, < 
does not increase by a multiple of 2ir in a period, and the motion as 
a whole is not periodic. 

The case of a body with an axis of symmetry. 

When the momental ellipsoid at the fixed point has an axis 
of symmetry, two of the throe moments of inertia A , B, C become 
equal to one another. This will be the case if the body is a solid 
of revolution of uniform density, but all that is actually required 
is the symmetry of the momental ellipsoid. The motion of the 
body under no forces is then greatly simplified. In fact, the 
simplification is so great that it is easier to discuss the problem 
afresh, rather than to apply the formulas of the general case. 
The motion can be determined, both qualitatively and quantita- 
tively, by the method of Poinsot. 

Let us denote the principal moments of inertia at by A and 
C, C being the axial moment of inertia and A the transverse 
moment of inertia. (This means that C is the moment of inertia 
about the axis of symmetry and A the moment of inertia about 
any perpendicular line through 0.) The cases A > C and 
A < C differ in some respects, but for the present we may treat 
them together. 

The Poinsot ellipsoid is of revolution. Since its center is 
fixed and it rolls on the invariable plane, the following facts are 
obvious : 

(i) The body cone and the space cone are both right circular 
cones. 

(ii) The angular velocity vector is of constant magnitude 
(w = OQ) and makes a constant angle with the invariable line OP 
(cf. Fig. 145). 

(iii) The invariable line, the angular velocity vector, and the 
axis of symmetry are coplanar at every instant. 

(iv) The axis of symmetry makes a constant angle (a) with the 
angular velocity vector and a constant angle (/3) with the invari- 
able line. 

To get a clear idea of the behavior of the body, let us start 
at the instant t = with the body in some definite position and 

'with some definite angular velocity <o, say OQo. Let ORo be the 
initial position of the axis of symmetry. Then a. = 



426 



MECHANICS IN SPACE 



[SEC. 14.1 



Let OSo be perpendicular to ORo in the plane RoOQo. We 
resolve o> along ORo and OS , obtaining components o> cos a 
and o) sin a. Since these are principal axes, the angular momen- 
tum vector h has components Ceo cos a along ORo and Av> sin a 
along 0$oJ it has, of course, no component perpendicular to the 
plane RoOQo. We are now in a position to construct h, and hence 
the invariable line OP. The angle (= R Q OP) is given by 



(14.122) 



tan 8 = 77 tan a. 



We have now to distinguish two cases, as shown in Figs. 146a 
and 6. 



N 









FIG. 146.- (a) The case where A > C. (b) The case where A < C. 



CASE (a): A > C (as in the case of a rod). Here /3 > a; the 
angular velocity vector lies between the axis of symmetry and the 
invariable line. 

CASE (b): A < C (as in the case of a flat disk). Here j3 < ; 
the invariable line lies between the axis of symmetry and the 
angular velocity vector. 

We have spoken of the instant t = 0. But a similar construc- 
tion may be made at any instant, and, as we have seen, the 
angles a and 0, and the magnitude of the angular velocity co are 
constants. So the figures we have constructed represent the 
state of affairs at any instant, the plane containing the figure 
rotating about the invariable line OP. This rotation is due to an 
angular velocity o> of constant magnitude, inclined to OP at a 
constant angle; hence the plane containing the axis of symmetry 
and the angular velocity vector rotates about OP with a constant 
angular velocity. We shall denote this angular velocity by ft. 

There is one more constant of importance. It is the angular 
velocity of the instantaneous axis about the axis of symmetry, as 



SEC. 14.1] 



MOTION OF A RIGID BODY 



427 



judged by an observer moving with the body. We shall denote 
this angular velocity by n. 

We have, in all, the following constants: 

a, 0, w, ft, n. 

Of these, a and co are determined by initial conditions; /3 is given 
by (14.122). We shall now set up equations to find and n, and 
at the same time get a clear picture of the motion by considering 
the space and body cones (Figs. 147a and 6). OP is the invari- 
able line, OQ the instantaneous axis, OR the axis of symmetry, 
and QN, QM are drawn perpendicular to OP, OR, respectively. 
In each case the space cone is fixed, and the body cone rolls on 



iace-Cone 



r-Cone 




Space-Gone 



O 

(a) (6) 

FIG. 147. (a) The case where A > C. (6) The case where A < C. 

it. This motion is easy to follow in Fig. 147a. The motion in 
Fig. 1476 may be understood by thinking of what the motion looks 
like from above, or by making a simple model out of thick paper 
and working the cones through the fingers in order to reproduce 
the condition of rolling. 

CASE (a) : A > C. The line OR turns about OP with angular 
velocity ft. Thus, in time dt, the point M receives a displacement 

OM sin ft ft dt = cos a sin OQ ft dt, 

perpendicular to the plane POR. But M is a point fixed in the 
body cone, which is turning about OQ with angular velocity w. 
Hence the displacement of M is also 

QM cos a o) dt = cos a sin a OQ o> dt, 



428 MECHANICS IN SPACE [Sue. 14.1 

since QM cos a is the perpendicular distance of M from OQ. 
Equating the two expressions, we obtain 



or, by (14.122), 

__ 

(14.124) ft = co ./sin 2 a + -p cos 2 a. 

To find n, we note that, in time dt, Q travels a distance QNtt dt 
on the space cone. But, from the definition of n, in the same 
time Q travels a distance QMn dt on the body cone. From the 
condition of rolling, these distances are equal to one another, and 
so 

QMn = QNil, 
or 

(14.125) n = Q 5L<?JL!> = sin <? ~ >. 

sin a sin /3 

In terms of the basic constants, we have 
(14.126a) n = A ~ C co cos a. 

The sense of this rotation n is obviously retrograde, when com- 
pared with co. 

CASE (b): A < C. The reasoning in this case follows the 
same lines, and we get the same formula (14.124) for ft, while 

(14.1266) n = j co cos a. 

The sense of this rotation n is direct, when compared with co. 

In the case of the earth, which is slightly flattened from the 
spherical form, we have A < C, and the ratio (C A) /A is 
small. Thus, case (6) applies, but in an extreme form, since the 
instantaneous axis is close to the axis of symmetry and a is 
small. The angular velocity n represents the rate at which the 
celestial pole, or axis of rotation of the earth, moves round the 
earth's axis of symmetry. For the period, we have approximately 



SEC. 14.2] MOTION OF A RIGID BODY 429 

If the sidereal day is taken as unit of time, then 27r/co = 1, and 
calculation gives the value 305 for the period. This prediction is 
in poor agreement with observation; for though a rotation of this 
sort is observed, its period is about 440 days. * The model used 
(a rigid body) proves at fault here, on account of the elasticity 
of the earth. 

Exercise. A circular disk is mounted so that it can turn freely about its 
center. Its angular velocity vector makes an angle of 45 with its plane 
and has a magnitude of 20 revolutions per second. Make a rough sketch 
of the space and body cones, and find J2 and n. 

14.2. THE SPINNING TOP 

The spinning top is the most familiar example of a gyroscopic 
system. The word "gyroscope" was invented to denote an 
instrument in which the earth's rotation produced an effect which 
could be observed. But the word "gyroscope" (or "gyrostat") 
is now used for any system in which a rapidly rotating body 
is so mounted that it may change the direction of its angular 
velocity vector. * 

Why does a spinning top not fall down? How does its rapid 
rotation render it apparently immune to the force of gravity, 
which makes non-spinning bodies fall? It is difficult to give a 
simple answer to this question. The only way to explain the 
phenomenon is to construct the mathematical theory of the top. 
For our purposes, we shall understand a "top" to mean a rigid 
body with an axis of symmetry, acted on by the force of gravity. 
A point on the axis of symmetry is fixed. Thus we idealize the 
ordinary top by supposing it to terminate in a sharp point (or 
vertex) and to spin on a floor rough enough to prevent slipping. 

Steady precession of a top. 

The motion of any rigid body with a fixed point satisfies the 
equation 

(14.201) h= G, 

where h is the angular momentum about and G the moment 
of the external forces about 0. In most dynamical problems, we 

* Of . H. N. Russel, R. S. Dugan, and J. Q. Stewart, Astronomy (Ginn and 
Company, Boston, 1945), Vol. I, pp. 118, 131-132. The section of the body 
cone by the earth's surface is a circle with a diameter of about 26 feet. 



430 



MECHANICS IN SPACE 



[SEC. 14.2 




mgK 



FIQ. 148. Vector diagram for top 
in steady precession. 



think of the forces as given and the motion as unknown; in that 
case, G is given and h is to be found. But we can look at (14.201) 
the other way round. We may regard the motion as prescribed, 

so that h is known as a vec- 
tor function of the time. Then 
(14.201) shows directly the mo- 
ment G which must be applied 
to the body in order to give this 
motion. 

Let us now describe a simple 
motion of a top, called steady 
precession, and inquire what forces 
must act on the top in order that 
this motion may take place. 

In steady precession, the axis 
of symmetry of the top describes with constant angular velocity a 
right circular cone with the vertical for axis. At the same time 
the top spins about its axis of symmetry with constant angular 
velocity. 

We shall use the following notation (Fig. 148): 

a = distance of mass center D from fixed vertex 0, 
m = mass of top, 

A = transverse moment of inertia at 0, 
C = axial moment of inertia at 0, 
K = unit vector directed vertically upward, 
(i, j, k) = unit orthogonal triad, with k along OD and i in the 

plane of k and K, 

= inclination of OD to the vertical. 
We note that 



(14.202) 



K = sin 6 i + cos 6 k. 



The angular velocity vector <o of the top lies in the plane 
(k, K). It can be resolved along i and k; we shall write 

(14.203) ii + 5k 

and call s the spin of the top. The velocity of the point D is 

6> X ak = (ii + sk) X ak = coiaj. 
We understand by the precession p the angular velocity with 



SEC. 14.2] MOTION OF A RIGID BODY 431 

which OD rotates about K. The velocity of D is then 

pK X ak = pa sin j. 

Equating the two expressions for the velocity of D, we have 
(14.204) coi = p sin 6. 

In the steady precession 0, s, and p are constants. 
The angular momentum is 



(14.205) h = Awt + Csk 

= Ap sin 0i + Csk. 

This vector lies in the plane (k, K), and rotates rigidly with it. 
Thus h is the velocity of a point with position vector h in a rigid 
body which turns with angular velocity />K. Therefore, 

(14.206) h - pK X h 

= p (sin 6 i + cos k) X (Ap sin 6 i + Csk) 
= p sin 6(Ap cos 6 Cs)j. 

The steady precession takes place, with assigned values of 
0, p, and s, provided that the moment about of all forces 
(including gravity) is 

(14.207) G = p sin B(Ap cos - Cs)j. 

Now the weight of the top is a force mgK at D and so has a 
moment 

ak X ( mgK) = mga sin 6 j 

about 0. If this is equal to G, as given by (14.207), no force 
other than the weight of the top is required to maintain the 
motion. Thus the steady precession takes place under gravity 
alone if 

(14.208) p(Cs - Ap cos 6) = mga. 

This is a single equation connecting the three constants 
6, p, s. There is, therefore, a doubly infinite set of steady 
precessions corresponding to arbitrary values of two out of the 
three constants. It is not, however, possible to assign com- 
pletely arbitrary values of two of the constants; these values 
must be such that (14.208) yields a real value for the third 
constant. 



432 



MECHANICS IN SPACE 



[SEC. 14.2 
In 



If we see a top spinning, 8 and p are easy to observe, 
terms of them, s is given by 

(14.209) . = V? + 4~L?. 

We note that, if the precession is small, the spin is great and 
is given approximately by 



(14.210) 



mga 
~Cp' 



This is a very simple and useful formula. 

Exercise. A disk, 6 inches in diameter, is mounted on the end of a light 
rod 1 inch long and spins rapidly. It processes once in 15 seconds. Find 
approximately the spin, in revolutions per second, and the velocity of a 
point on the edge of the disk. 

General motion of a top. 

To discuss the general motion of a top, we shall use the same 
notation for the constants of the top as that used above. 




FIQ. 149. Vector diagram for top in general motion. 

Let I, J, K (Fig. 149) be a fixed orthogonal triad, K being 
directed vertically upward. Let i, j, k be an orthogonal triad, 
with k pointing along OD, the axis of symmetry of the top, and 
i coplanar with k and K; thus j is horizontal. The triad i, j, k 
is fixed neither in space nor in the top, but k is fixed in the top. 

Let 6, <t> be the usual polar angles of k relative to the fixed 
triad. Variations in are referred to as nutation, and variations 
in <j> as precession. 

Let 

(14.211) <> = d>ii -f- GJ2J ~~h coak 



SEC. 14.2] MOTION OF A RIGID BODY 433 

be the angular velocity of the top, and 

(14.212) a = ftii + 2 j + Qsk 

the angular velocity of the triad i, j, k. It is easy to see that 

(14.213) Q! = sin <, Q, = -0, U 3 = cos 6 <. 

Now the relative motion of the top and the triad i, j, k consists 
only of a rotation about k. Hence, 

(14.214) coi = fli = sin B <, o> 2 = Q 2 = 0. 
The angular momentum is 

(14.215) h - Awii + Ao> 2 j + Cwjc, 
and its rate of change is, by (12.306), 

(14.216) h = Ahi + Aw 2 j + Cwak + O X h. 
The moment about of the weight of the top is 

(14.217) G = ak X (-nigK) = -mga sin j. 

The motion of the top satisfies the fundamental equation 

(14.218) h = G. 

When we substitute the expressions given above, this vector 
equation gives three scalar equations for 0, <, and w 3 . However, 
an indirect method of attack proves simpler, and we shall make 
direct use only of the third component of (14.218). 
The component of (14.218) in the direction of k gives 

Ctos = 0, 

since, by (14.214) arid (14.215), 12 X h has no component in the 
direction k. Hence 

(14.219) <a t = s, 

a constant; the spin of the top is a constant. Further, since the 
weight of the top has no moment about K, the component of 
angular momentum in this fixed direction is constant, and so 

(14.220) h K = a, 

a constant. By (14.214) and (14.215), this gives immediately 

(14.221) Aj> sin 2 + Cs cos 6 = a, 



434 MECHANICS IN SPACE [SEC. 14.2 

since K = sin 6 i + cos 6 k. Finally, we have the equation of 
energy 

(14.222) T + V = E, 
or 

(14.223) iA(wf + wf) + iCw| + mga cos 9 = E, 

E being a constant. Substitution from (14.214) and (14.219) 
gives 

(14.224) A(6 2 + & sin 2 6) + Cs 2 = 2(E - wgfa cos 0). 



We have in (14.221) and (14.224) two equations to determine 
6 and < as functions of the time. 

It is convenient to write Cs = /?. Then our two equations 
read 

A<j> sin 2 = a p cos 6 

* 2 sin 2 61) + = 2(# - m0 cos 0). 



The plan is now obvious. We are to substitute for < in the 
second equation from the first; this will give a differential equa- 
tion for 6. When this is solved, the first equation, will give < 
by a quadrature. 

Let us put x cos 0. On multiplying tho second equation 
in (14.225) by sin 2 6 and substituting for <, we obtain for x the 
differential equation 



(14.226) . 

L -rx- j v/ 

= 2(E - mgax)(l - 
This equation may be written 

(14.227) **=/(*), 
where 

(14.228) /(*) = I 

This is a cubic in x, and, by the same form of argument as that 
used in Sec. 13.3 for the spherical pendulum, we see that it has a 



SEC. 14.2] 



MOTION OF A RIGID BODY 



435 



graph of the general form shown in Fig. 150; the function f(x) 
has three real zeros x\, # 2 , # 3 , such that 

1 < Xi < x z < 1 < 3. 

(In special cases, we may have one or more signs of equality 
instead of inequality.) Thus /(x) may be written 



(14.229) /(*) = 



(x - 



Again by the same argument as in Sec. 13.3, the solution of 
(14.227) is 

(14.230) cos = x = xi + (x* - xi) sn 2 \p(t - / )1, 

f(x) 




Fia. 150. Graph of f(x) for a top in general motion. 

where p and the modulus k of the elliptic function are given by 
. _ mga(x> - *,) , 2 _ JT, -Xi 



(14.231) 



2A 



3 ~ 



The constants x\, x z , x^ are functions of the constants occurring 
in (14.228), i.e., the constants of the top and a, 0, E; the latter 
are known when the initial position and angular velocity of the 
top are given. 

The complete solution for the motion of the axis of the top is 
given by (14.230) and 



a fa 



(14.232) 



Since x is known as a function of t y this last equation gives 
by a quadrature. 

This analytic solution does not immediately give a clear 
idea of the way in which the top behaves. However, we can 



436 MECHANICS IN SPACE [SEC. 14.2 

construct the essential features of the motion, by fixing our 
attention on the intersection of the axis of the top with a unit 
sphere having its center at 0. It is interesting to compare the 
motion of this point with the motion of a spherical pendulum. 

In the first place, it is clear from (14.230) that the representa- 
tive point on the unit sphere oscillates between two levels 
B = 0i and = 2 , given by 

cos 0i = #1, cos 2 = #2. 

This behavior is like that of the spherical pendulum; but while 
the mean level for the spherical pendulum must lie below the 
center of sphere, that is no longer necessarily true for the top. 
There is, however, a more striking difference; in the case of the 




a b 

FIG. 151. Motion of the axis of a top. (a) without loops, (b) with loops. 

top, we may have loops on the curve. The absence of loops, 
as in Fig. 151a, or the presence of loops, as in Fig. 1516, depends 
on the way in which the motion is started, i.e., on the values of 
the constants a, /?, E. The criterion for the existence of a loop 
is that <j> should sometimes increase and sometimes decrease, 
and the condition for this is that <j> should vanish during the 
motion. By (14.232), < = when x a//3; since x oscillates 
between x\ and # 2 , it is just a question as to whether a/0 lies 
in this range of oscillation. If it lies in the range, there are 
loops; if not, there are no loops. 

Cuspidal motion of a top. 

A particularly interesting case arises when the top is spinning 
with its axis fixed in position and then released. It starts to 
fall but recovers and rises to its former height, repeating this 
process over and over again. 

This case can be discussed in terms of the theory just developed. 
The behavior of a top depends essentially on the cubic f(x) of 
(14.228), and to it we must direct our attention. 



SEC. 14.2] MOTION OF A RIGID BODY 437 

First, let us note that initially x x (say), x = 0, and $ = 0. 
Hence, by (14.232), a = #c ; also, by (14.226), 



Substitution in (14.228) gives 

(14.233) f(x) = ^ ( Xo - z)(l - * 2 ) ~ J 2 (*o - *) 2 . 

Obviously, one zero of f(x) is x = #o; but is this zero #1 or 
Differentiation gives, for x XQ, 

f(xo ) = _ 



Since this value is negative, it is clear from Fig. 150 that x = z 2 , 
not XL 

The oscillation of x is from xi to #o (or x 2 ), where Xi is the 
smallest zero of /(x). Putting 



(14.234) 
v x 



we see that the three zeros of f(x) are 



a: 2 = so, 

Z 3 = X + VX 2 - 2Xx + 1. 

Thus the axis of the top falls down from an inclination 0o 
(where cos 0o = #o) to an inclination 0i, where 



(14.236) cos 0i = xi = X - VX 2 - 2Xx + 1. 

Then it starts to rise again and swings up to 6 , where the 
axis is again instantaneously at rest. 

If X is large, i.e., if the spin is great, binomial expansion of the 
radical in (14.236) gives approximately 

sin 2 0o 
cos 0i = cos H\ 

The difference 0i is small, and so we may use the approxima- 
tion 

cos 0i - cos = ~ (0i 0o) sin 8 . 



438 MECHANICS IN SPACE [Sue. 14.2 

The axis falls only through the small angle 
(14.237) 0i = ^!Tf a s i n 0> 

The differential equation of the path of the representative 
point on the unit sphere is 

dx x 4(1 #2) ^/J(x) 

Since /(x) vanishes like x XQ as x > o, it is clear that d<j>/dx = 
at the highest positions of the axis. Hence the path of the 
representative point meets the circle = 0o at right angles; the 
path has cusps at these points, directed upward. 

Stability of a sleeping top. 

Anyone who has seen a top spinning is familiar with the 
general nature of the motions discussed above. Sometimes 
the top executes a motion of steady procession, and sometimes tho 
more general motion in which the axis of the top nods up and 
down as it processes. A third typo of motion is often seen, 
in which the top spins with its axis vortical. The axis remains 
stationary and there is no apparent motion of the top as a whole- 
it is then said to be a sleeping top. A small disturbance of a 
sleeping top produces only a small oscillation when the spin 
is great; when tho spin has boon considerably reduced by fric- 
tional resistance, the top begins to wobble and ultimately falls 
down. We naturally ask: What is the critical value of the spin 
below which the motion of a sleeping top is unstable? 

The answer is found by examining the cubic f(x) given in 
(14.228). Since = = for a sleeping top, we have, by 
(14.221) and (14.224), 

a = = Cs, E = |Cs 2 + mga\ 
hence, (14.228) gives 

(14.238) 



We observe that x = 1 is a double zero of f(x). Two cases 
arise: either the third zero of f(x) is greater than unity, or it 
is less than unity. The forms of the graph of f(x) for these two 



SEC. 14.2] 



MOTION OF A RIGID BODY 



439 



cases are shown in Figs. 152a and 6; in terms of the notation 
used for the zeros of /(#), Fig. 152a shows the case where the 
third zero is x s , and Fig. 1526 the case where it is x\. 

When the top suffers a small disturbance, the graph of f(x) 
for the disturbed motion will not be the same as that for the 

f(x) 




f(x) 




FIQ. 162.- (a) Graph of f(x) for a stable sleeping top. (b) Graph of f(x) for 
an unstable sleeping top. In each case, the broken curve is the graph for dis- 
turbed motion. 

sleeping top. The difference will be small, however, since only 
small changes in the constants a, #, E can result from a small 
disturbance. The broken curves in Figs. 152a and b indicate the 
way in which the graphs of f(x) are modified by a small dis- 
turbance. In general, the three zeros of f(x) will become dis- 
tinct two of them must, of course, lie in the range ( 1, 1), 
and the third must exceed unity. 



440 MECHANICS IN SPACE [SBC. 14.2 

Since, in the disturbed motion, the value of x lies between 
the two smaller zeros of f(x), it is clear that one or other of the 
following descriptions applies: 

(i) The axis of the top, when disturbed, does not depart far 
from its original vertical position the motion is stable. This 
corresponds to Fig. 152a. 

(ii) The axis of the top falls to an inclination 0i, where 

cos 0i = #1 

the motion is unstable. This corresponds to Fig. 1 526. 

To find the critical value of the spin s, it remains to distinguish 

the two cases analytically. 

The two types of curve are distinguished by the sign of /"(#) 
at x = 1; it is negative in Fig. 152a and positive in Fig. 1526. 
Differentiating (14.238), we find 



this is negative and the motion of a sleeping top is stable, if 



(14.239) s> > - 

In the limiting case s 2 = 4Amga/C 2 , it is easy to see that all 
three zeros of f(x) coincide at a; = 1, and the motion is stable. 

Exercise. Show that, for a motion of steady procession, the cubic f(x) 
has a double zero lying in the range ( 1, 1). Hence show that this type of 
steady motion is always stable. 

Stability of a spinning projectile. 

It is a well-known fact that an elongated projectile acquires 
stability from the spin imparted to it by the rifling in the gun. 
By this we mean that it does not turn broadside on to its direc- 
tion of motion when in flight, nor does it tumble as a nonspinning 
projectile often does. We shall now use our theory of the 
motion of a top in an attempt to explain this spin stabilization. 

In Sec. 6.2 we discussed the motion of a projectile in a resisting 
medium, the projectile being treated as a particle. The problem 
becomes much more complicated when the projectile is treated 
as a rigid solid of revolution, subject to gravity and to the aero- 
dynamic force system due to the pressure of the air. We cannot 



SBC. 14.3J MOTION OF A RIGID BODY 441 

enter here into this general problem; instead, following the older 
writers on ballistics, we shall make some drastic simplifications. 
Thus, we shall assume that the aerodynamic force system is 
equipollent to a single force (the drag) with fixed direction and 
constant magnitude (R), intersecting the axis of the projectile 
at a fixed point (the center of pressure). We can now use Fig. 
149, with suitable changes, for the discussion of the motion of the 
projectile relative to its mass center. Let be the mass center, 
K a fixed unit vector opposed to the direction of the drag, and k 
a unit vector along the axis of the projectile. The point D is 
taken to be the center of pressure; and the force at D, i.e., mgK 
in the case of the top, is now to be replaced by RK. The 
weight of the projectile acts through 0, and so the total moment 
about is due to the aerodynamic forces alone; it is 

(14.240) G = -ftasin 0j, 

where a is the distance of the center of pressure in front of the 
mass center and 6 the tingle between the axis of the projectile 
and the direction of the drag reversed. 

Since the fundamental equation (12.209) for motion relative 
to the mass center is of the same form as (14.218) and the expres- 
sion (14.240) for G is of the same form as (14.217) with the con- 
stant mga changed to Ra, we can apply to the motion of the 
projectile the same mathematical analysis as we applied to the 
motion of the top. In particular, if the projectile is moving 
along its axis, the fixed direction of the drag will also lie on this 
line, and we have what is essentially a sleeping top. Then 
(14.239) yields the condition for stability of the spinning pro- 
jectile, i.e., the condition that the projectile, if slightly disturbed, 
will not develop large oscillations. This condition is 

(14.241) * > ^, 

where s is the spin of the projectile, A the transverse moment of 
inertia at (i.e., at the mass center), and C the axial moment of 
inertia. 

14.3. GYROSCOPES 

The stability of a gyroscope. 

Let us suppose that a gyroscope (i.e., a rigid body with an 
axis of symmetry) is mounted in a Cardan's suspension (Fig. 144), 



442 MECHANICS IN SPACE [Sue. 14.3 

so that its mass center is fixed. It is set spinning about its axis 
of symmetry with a great angular velocity s. Now let an 
impulsive couple G be applied to the gyroscope. The instan- 
taneous change in angular momentum is [cf. (12.502)] 

(14.301) Ah = G. 

We have then a vector diagram as in Fig. 153. The vector OA is 
h, the angular momentum before the impulsive couple was 

B 




Fio. 153. Change in angular momentum due to an impulsive couple. 

applied. It is made long, because s (and consequently /?,) is 

> > 

assumed to be large. The vector AB is Ah, and OB represents 

the final angular momentum. It is clear that the angle AOB 
is small; in fact, it tends to zero as s tends to infinity. Thus, 
the application of an impulsive couple to a rapidly spinning 
gyroscope makes only a small change in the direction of the 
angular momentum vector. It is easily seen that the correspond- 
ing change in the direction of the angular velocity vector is also 
small. 

This simple result illustrates the stability which a rapid 
rotation imparts to a body. The gyroscope shows, as it were, 
an unwillingness to alter the direction of its axis. When it does 
yield, it does so in a manner which continues to cause surprise 
even to those familiar with the theory. 

The gyroscopic couple. 

In Fig. 154a, is a fixed point on the axis of a gyroscope, 
and j a unit vector fixed in space. As pointed out earlier, any 
motion can be produced, provided suitable forces are applied. 
Let us demand that the gyroscope shall spin with constant 
angular speed s about its axis, and at the same time that the axis 
shall turn (or process) with constant angular speed p in the plane 
perpendicular to j. If k is a unit vector along the axis of the 
gyroscope and i completes the triad, then the angular velocity 
of the gyroscope is 
(14.302) <o = pj + sk, 



SEC. 14.3] MOTION OF A RIGID BODY 443 

and the triad (i, j, k) has an angular velocity 

(14.303) 11 = pj. 

If A and C are, respectively, the transverse and axial moments 
of inertia, the angular momentum is 

(14.304) h = Apj + Csk, 
and its rate of change is 

(14.305) h = a X h - Cspi. 

Thus the gyroscopic couple G required to maintain this motion is 

(14.306) G - Cspi, 

that is, a couple of magnitude Csp, produced by a pair of forces 
in the rotating plane of j and k. 



Cs 




Ap 



(a) 

FIQ. 154. (a) Angular momentum diagram for a processing gyroscope. 
(b) Relations between couple, precession, and spin. 

The relations between the couple, the precession, and the spin 
are shown in Fig. 1546. It is more interesting here not to show 
the vectors in the usual way, but to represent them by arcs in 
the planes perpendicular to them. The curious fact, hard to 
understand intuitively, is that the plane of the couple does not 
coincide with the plane of the precession, but is perpendicular 
to it. Instead of yielding to the couple, the axis of the gyroscope 
turns at right angles to the plane of the couple. 

It is evident from (14.306) that when the gyroscope spins 
rapidly, a very great couple is required to produce even a mod- 
erate rate of precession. 

Although the diagram of Fig. 1546 shows only a simple gyro- 



444 



MECHANICS IN SPACE 



[SEC. 14 3 



scopic phenomenon, it is very useful from a practical standpoint. 
We note that the three quadrants form a single closed curve. 
As we traverse it in the sense indicated by the arrows, we cover 
the following quadrants in order: 

Couple, 

Precession, 

Spin. 

This is easy to remember, since the letters C, P, S are in alpha- 
betical order. 

Example An airplane has a rotary engine, which rotates in a clockwise 
direction when viewed from behind. The airplane makes a left turn. Does 

the gyroscopic effect of the rotating engine tend 
to make the nose rise or fall? 

First we suppose that the pilot sots the 
rudder and elevator in such a way that the 
nose goes neither up nor down. The mass 
center of the engine describes a circular arc 
C as the airplane turns (Fig. 155). The 
angular velocity of the engine is made up of 
a large component along the tangent to (7 
and a small vertical component, due to the 
turning of the airplane as a whole. The 
quadrants of precession and spin are therefore 
as shown Hence, by the rule of alphabetical 
order, the couple quadrant comes down in 
front. To maintain the motion described, 
the pilot must set the rudder and elevator in 
such a way that aerodynamic forces, acting 
on them, produce the required couple. 

If the pilot flies the airplane with a rotary 
engine in the same way as he would fly a similar an plane with a stationary 
engine, the couple required to maintain the steady flight in a horizontal circle 
will not be present. Since the couple is one which tends to depress the nose, 
in its absence the nose will rise. What will happen after the initial lift takes 
place is a complicated question, not covered by the present simple theory. 

The gyrocompass. 

If the spinning of the earth on its axis were much faster than 
it actually is, it would be a simple matter to devise a mechanism 
by which the true north could be found on a ship at sea. How- 
ever, the earth's rotation is so slow that an apparatus of great 
delicacy is required, in order that a minute effect may not be 
wiped out by frictional resistances. The modern gyrocompass 




FIG. 155. Airplane turning. 



SEC. 14.3] 



MOTION OF A RIGID BODY 



445 




is such a piece of apparatus. The simple system which we shall 
discuss is much less elaborate than the gyrocompass as it is 
actually constructed.* However, the basic fact that a spinning 
gyroscope enables us to find the north is demonstrated by a 
discussion of an ideally simple gyrocompass. 

In Fig. 156, PQ is part of the earth's axis, drawn from south 
to north. The point is on the earth's surface at latitude A. 
Thus the horizontal line, drawn due north from 0, is inclined 
to the earth's axis at an angle X; this line is OQ in the diagram. 
The unit vector K is parallel to PQ. 

A gyroscope is mounted in a Cardan's 
suspension (Fig. 144) so that its mass center 
lies at 0. But it is not left froo to turn 
about 0} one pair of the bearings in the 
suspension is locked, so that tho axis of the 
gyroscope can move only in tho horizontal 
plane at 0. The unit vector k lies along 
the axis of the gyroscope, making with OQ 
a variable angle 8; i is perpendicular to k 
in the horizontal plane, and j is vertical 
(i.e., coplamir withPQ, OQ uml porpcndi- 
cular to OQ). Tho gyroscope itsolf is not 
shown in the diagram; the curve is a unit circle in the horizontal 
plane. 

The angular velocity of the triad (i, j, k) is mado up of the 
angular velocity of the earth, which we may write 2K, and an 
angular velocity due to change in 0, in fact, 0j. Since 

K = sin 6 cos X i + sin X j + cos 6 cos X k, 
the angular velocity of the triad is 

(14.307) o>' = - ft sin cos X i + (6 + sin X)j 

+ ft cos 6 cos X k. 

The angular velocity o> of the gyroscope differs from this only in 

* Cf. H. Lamb, Higher Mechanics (Cambridge University Press, 1929), 
p. 144; R. F. Deimol, Mechanics of the Gyroscope (The Macmillan Com- 
pany, New York, 1929); A. L. Rawlings, The Theory of the Gyroscopic 
Compass (The Macmillan Company, New York, 1929); E. S. Ferry, Applied 
Gyrodynamics (John Wiley & Sons, New York, 1932). 



P 

FIG. 156. Vector dia- 
gram for gyrocompass. 



446 MECHANICS IN SPACE [SBC. 14.3 

the third component; thus, 

(14.308) o> = - ft sin cos X i + ( 6+ ft sin X)j + sk, 

where s is the axial spin, including a component of the earth's 
rotation. The angular momentum is 

(14.309) h = -A 12 sin cos X i + A (6 + Q sin X)j + Csk, 

where ^4 and C are, respectively, the transverse and axial moments 
of inertia. 

To keep the axis of the gyroscope in the horizontal plane, 
the bearings of the suspension must exert a couple G on it. 
Since the bearings are assumed to be smooth, no work is done 
by this couple in rotations about either j or k. Hence, G is 
perpendicular to these vectors, and we may write 

(14.310) G = Gi, 

where G may be either positive or negative. 
The fundamental equation h = G gives 

(14.311) -AQcoa B cos X 6i + A 8j + 6 Y .sk + ' X h = Gi. 

We now pick out the j and k components of this equation and so 
obtain two scalar equations for s and 0: 



, . I ^ ~^~ Cstt cos X sin 9 AiT 2 sin 9 cos 6 cos 2 X = 0, 

(14-.ol.ttJ 1 yry. ,- 

The second equation shows that the spin s is constant. With 
12 2 neglected, the first equation may be written 

(14,313) S + n* sin (9 = 0, 

where 

/CsiTcos X 
n -i 



(14.314) n = J- 



Now (14.313) is the equation of motion of a simple pendulum. 
It is possible for the axis of the gyroscope to go right round 
the horizontal circle, but if the initial values of 6 and 6 are small, 
the motion will be oscillatory. The important fact is this: 
The axis of the gyroscope oscillates symmetrically about the direction 
6 = 0. Hence, by bisecting the angle of swing, we may find 
the north. The gyroscope therefore acts as a gyrocompass. 



SEC. 14.4] MOTION OF A RIGID BODY 447 

indicating the true north; the magnetic compass, of course, 
indicates the magnetic north. 

For small oscillations, the periodic time of swing for the gyro- 
compass is 

(14.315) r = = 2w LA v 

v ' n \Csl2cosX 

Since 12 is so small (1 revolution per sidereal day = 2r/86,164 
radians per second), the spin of the gyroscope (s) must be given 
a large value in order to make r reasonably small. If X = ?r, 
the periodic time becomes infinite, and the gyrocompass fails to 
function; but that is only to be expected, for the points in ques- 
tion are the North and South Poles. 

Exercise Assuming the mass of the gyroscope concentrated in a thin 
ring, find the number of revolutions per second required for a periodic time 
of 10 seconds at latitude 45. 

14.4. GENERAL MOTION OF A RIGID BODY 

The general motion of a rigid body consists of (i) motion 
of the mass center and (ii) motion relative to the mass center. 
The equations governing these have been given in Sec. 12.4. 
But it would be wrong to suppose that the determination of the 
general motion always splits into two parts a problem in 
particle dynamics and a problem in the dynamics of a body with 
a fixed point. Constraints make the two problems interlock, 
and complications arise. We cannot give a general plan for 
the solution of all such problems but shall determine the motions 
of two systems as examples. 

The motion of a billiard ball. 

A billiard ball is struck by a cue. At time t = 0, we suppose 
that the ball is in contact with the table; its center has a horizon- 
tal velocity qo, and the ball has an angular velocity G>O. We wish 
to find the subsequent motion of the ball. 

If the table were perfectly smooth, the center of the ball would 
retain the velocity q , and the angular velocity w would also be 
retained. But we shall assume the table to be rough, with a 
coefficient of kinetic friction ju. 

At a general time t, the center has a horizontal velocity q, 
and the ball has an angular velocity <o. Let K be a unit vector 



448 



MECHANICS IN SPACE 



[SEC. 14.4 



drawn vertically upward (Fig. 157). The reaction of the table 
on the ball at the point of contact P may be written 



where R is the magnitude of the normal reaction and F the force 
of friction; F, of course, acts horizontally. The weight is mgK, 
where ra is the mass of the ball. Thus the equation of motion 
of the center is, by (12.410), 

(14.401) mq = F + (R - mg)JL. 

But, since the ball remains in contact 
with the table, the acceleration of the 
center has no vertical component. 
Thus R = mg, and we have 




(14.402) 



mq = F. 



FIG. 157. Ball slipping on 
a table. 



Since every axis at is a principal axis 
of inertia, the angular momentum about 
O is h = 7wA- 2 co, where k is the radius of 

gyration about a diameter. Thus the equation for motion relative 

to is, by (12.411), 

(14.403) roA-% = -aK X (F + KK) = -aK X F, 

where a is the radius of the ball. 

In the vector equations (14.402) and (14.403), there are 
actually five scalar equations. There are seven unknowns, viz., 
two components of q, throe components of <o, and two components 
of F. Thus, two more equations are required; they are furnished 
by the law of kinetic friction, as long as there is slipping betweon 
the ball and the table. This law tells us that F acts in a direction 
opposite to the velocity of the particle of the ball at P, and that 

(14.404) F = nR 
Thus, 

(14.405) F - - 



where q' is the velocity of the particle at P, given by 



(14.406) 



q' = 



c* X (-aK). 



SEC. 14.4] MOTION OF A RIGID BODY 449 

Hence, using (14.402) and (14.403), we get 
(14.407) mq' = F + ~ (K X F) X K 



since K F = 0. Thus, by (14.405) and (14.407), the derivative 
of q' has a direction opposed to q'. This implies that, as long 
as slipping is taking place, the vector q' has a fixed direction. 
Let I be a unit vector in this fixed direction. Then 

(14.408) q' = 01, F = -/impl, 

and, by (14.407), the magnitude of q' changes according to the 
equation 

(14.409) <?'= - 
Hence, 



where q' Q is the magnitude of qj, the initial velocity of slipping, 
viz., 

(14.411) qj = q () - a<o X K. 
By (14.410), we shall have </ = when 

(14.412) t = -^ - -r-i^; 
v ' M a 2 + A- 2 ' 

at this instant slipping ceases. It is easy to see that, once slip- 
ping ceases, the motion becomes a simple rolling in a straight line 
at constant speed, for (14.402) and (14.403) are satisfied by 
constant values of q and w, with F = 0. 

There is a point of interest in connection with the motion of 
the center before slipping ceases. By (14.402) and (14.408), 
we have 

(14.413) 4 = -M0I- 

This means that the acceleration of the center of the ball is 
constant in direction and magnitude, and so the center describes 
a parabolic path as long as slipping persists. 



450 MECHANICS IN SPACE [Sac. 14.4 

The above results hold for any ball in which there is a spheri- 
cally symmetric distribution of matter. If the ball is solid and 
homogeneous, we put 2 = fa 2 . 

The motion of a rolling disk. 

Everyone knows that a child's hoop, or a rolling coin, acquires 
stability from its motion. If the hoop or coin rolls slowly, it will 
start to wobble violently, but if it rolls fast, it can pass over small 
obstacles without being upset. We shall now discuss such 
motions, idealizing for simplicity to the case where the body has 
a sharp rolling edge. We can treat the hoop and the coin (or 
indeed any body with a sharp circular 
edge, possessing an axis and a plane of 
symmetry) in a single argument by using 
general symbols for moments of inertia. 
For purposes of reference, however, we 
shall uso the word "disk." 

Figure 1 58 shows the disk in a general 
position; P is the point of contact with 
the ground, which we shall suppose rough 
FIG. 158. Dibk rolling on enough to prevent slipping. Let 6 be 
a plane. ^ mc ii na tion of the plane of the disk 

to the vertical, and $ the angle between a fixed horizontal 
direction and the tangent to the disk at P. Let (i, j, k) be a unit 
orthogonal triad, k being perpendicular to the disk at its center 
and i lying along the radius toward P] j is therefore horizontal 
and lios in the piano of the disk. 

For the velocity of the center and the anguhir velocity of the 
disk, we may write 

q = ui + v] + wk, o> = o>ii + oj 2 j + wsk. 

These two vectors are not independent, because the particle at 
P is instantaneously at rest. This gives the condition 

q + o> X ai = 0, 

where a is the radius of the disk; or, in scalar form, 
(14.414) u = 0, v + aojs = 0, w aui = 0. 

These equations determine q when o> is known. 

Now the angular velocity Q of the triad arises solely from 




SEC. 14.4] MOTION OF A RIGID BODY 451 

changes in 6 and <. The former gives an angular velocity 
0j, and the latter an angular velocity <j> about OQ, the vertical 
through 0. Thus, 

(14.415) a = - cos fa - 6j + sin 6 <k. 

But the angular velocities of the disk and the triad differ only 
in the k component. Therefore 

(14.416) coi = - cos B <l>, o) 3 = -0. 
For the reaction of the ground, we write 

(14.417) R = flii + # 2 j + ft 3 k. 

By (12.410) and (12.411), the two vector equations of motion 
are 

( mq = R + ?rc<7(cos i sin 6 k), 

} h = ai X R, 

where m is the mass of the disk and h is the angular momentum 
about 0; we have 

h = Acoii + .<4co 2 j + Ow 3 k, 

A and C being transverse and axial moments of inertia at O. 
Since, by (12.306), 

(14.419) q = -wi + j + ibk + a X q, 

the first of (14.418) gives the throe scalar equations 

(m(u Ow sin <<>) = li\ -f- mg cos 0, 
m(v + sin 6 fat -f cos <M = 72 a , 
m(?/? cos fa) \ 611) 7?3 ing sin 0. 

By (14.414) and (14.116), we eliminate ?/, *>, i/J and obtain 

(ma(0 2 + sin <w 3 ) = /?i + ^.<7 os 0, 
-?/i<z(w3 + cos 00<^) - K 2 , 
ma($ cos 0a? 3 ) = /? 3 7??^r sin 0. 

Turning now to the second of (14.418), we have 
(14.422) h = Aciii + A^j + 6'wjc + ft X h, 

and so we get the three scalar equations 

(Awi <70co 3 A sin ^w 2 = 0, 
Aw 2 + A sin <wi + (7 cos <co 3 = -a/^ 3 , 
Cd) 3 A cos <a> 2 + A 0a)i = 



452 MECHANICS IN SPACE [SEC. 14.4 

By (14.410), these become 



(14.424) 



A -~ (cos B <) + C0o>3 - A sin 0< = 0, 



Associating these equations with (14.421), we have six equations 
for the six unknowns 6, <, cos, Ri t Rz, Rz- They are the equations 
(12.412) applied to our special problem. 

Before proceeding to discuss the stability of the disk rolling 
straight ahead, let us consider simple steady motjons satisfying 
the equations (14.421) and(14.424). 

The most obvious solution is 



(14.425) f'" ' * 
v I R l = -mg, 



= constant, 0)3 = constant, 



This is the straight-ahead motion, in which the plane of the disk 
is vertical. Another simple motion is given by 

(14.426) = constant, <# = constant, co 3 = constant. 
The corresponding reactions are, by (14.421), 

I HI = m(a sin B <j>& 3 g cos 0), 
#2 = 0, 
/^s = m(a cos 6 ^>oj 3 + g sin 0). 

When we substitute in (14.424), the satisfaction of these equa- 
tions requires 

(14.428) (C + ma 2 ) cos 6 <a> 3 + mga sin 6 = A sin cos e < 2 ; 

this condition must be satisfied by the constant values of 0, <, 
w 3 , in order that the steady motion may exist. It is, of course, 
a rolling in a circular path. 

Exercise. If, in the motion given by (14.426), and < are small, show that 
the time taken to complete the circular path is approximately 



Let us now discuss the stability of the rolling disk. We 
suppose the disk to be slightly disturbed from the steady motion 
given by (14.425). In the disturbed state the following quanti- 



SEC. 14.5] MOTION OF A RIGID BODY 453 

ties are assumed to be small, since they vanish in the steady 
motion: 

(14.429) 0, 6, S, <, <, o> 3 , R 1 + mg, R Zl R 9 . 

With only first-order terms retained, (1 1.421) and (14.424) 
become 

(0 = #1 + nig, A<t> + (70o>3 = 0, 
mau* = -#,, AS - Cfa* = aR 9 , 
maS mafas /?a + mgO, Cu* aR>>. 



From the second and last equations we see that wa = constant. 
Elimination of < and R s from the other equations gives 

(14.431) A(A + ma*)8 + [C(C + ma 2 )o>S - Amga]0 = a, 

where a is a constant of integration. Obviously the condition 
for stability is 



(14.432) 



C(C + ma 2 ) 



14.6. SUMMARY OF APPLICATIONS IN DYNAMICS IN SPACE- 
MOTION OF A RIGID BODY 

I. Rigid body with fixed point under no forces. 

(a) For a general body, the motion is given by rolling the 
Poinsot ellipsoid on the invariable plane; there is an analytic 
solution in terms of elliptic functions. 

(b) For a body with an axis of symmetry, the Poinsot ellipsoid 
is of revolution, and the motion is given by rolling the right 
circular body cone on the right circular space cone at a constant 
rate. 

II. The spinning top. 

(a) Steady precession (p) with fast spin (s) : 



(14.501) s = -y- (approximately). 

op 

(b) General motion expressible in terms of elliptic functions. 

(c) Sleeping top stable if 

(14.502) s* > 



454 MECHANICS IN SPACE [Ex. XIV 

III. Gyroscopes. 

(a) A finite impulsive couple, applied to a fast-spinning 
gyroscope, alters the direction of the axis of rotation only slightly. 

(6) Gyroscopic couple G required to maintain precession p: 

(14.503) G = Csp. 

Couple > Precession Spin. 

(c) Gyrocompass: 



( 1 4.504) r = 27r T, 

v 7 



cos X 

IV. General motion of a rigid body. 

(a) Center of a slipping billiard ball describes a parabola. 

(b) Rolling disk is stable if its angular velocity w satisfies 



(14.505) o,* > - cT <f +^ 

EXERCISES XIV 

1. A circular disk, pivoted at its center, is set spinning with angular 
velocity o> about a line making an angle a with its axis. Find, in terms of 
w and a, the time taken by the axis of the disk to describe a cone in space. 

2. A gyroscope can turn freely about its mass center which is fixed. 
Initially, it is set spinning about its axis, which is then struck perpendicu- 
larly. Find the angular velocity immediately after impact in terms of the 
initial spin, the moments of inertri, the magnitude of the impulsive force, and 
its distance from 0. Draw a diagram showing the direction of the impulsive 
force, the angular velocity, and the angular momentum just after impact. 

3. The center of a square plate is fixed. If at a certain instant the 
angular velocity vector makes an angle of 30 with the normal to the plate, 
find the inclination of the angular momentum vector to the normal. 

4. A rigid body turn?? about a fixed point under the action of a single 
force F (in addition to the reaction at the fixed point). If the extremity 
of the angular momentum vector, drawn from O, lies in a fixed plane P 
throughout the motion, show that F intersects or is parallel to the perpen- 
dicular dropped from O on P. 

5. A heavy homogeneous right circular cone spins with its vertex fixed. 
The axis of the cone is 4 in. long, and the radius of the base is 2 in. The 
axis maintains a constant inclination to the vertical and completes a rotation 
about the vertical in 5 sec. Find approximately the number of revolutions 
per second of the cone about its axis. 

6. A lamina turns freely under no forces in three-dimensional motion 
about its mass center, which is fixed. Use Euler's. equations to prove that 
the component of angular velocity in the plane of the lamina is constant in 
magnitude. 



Ex. XI VJ MOTION OF A RIGID BODY 455 

7. A rigid body turns about a fixed point under no forces. The momen- 
tal ellipsoid at the fixed point ih of revolution, and the axial moment of 
inertia C is greater than the transverse moment of inertia A. Show that 
if a is the angle of inclination of the instantaneous axis to the axis of sym- 
metry, then the semiangle of the space cone is 

tun-* (C ~ 4Ltan_ 
C + A tan 2 a 

Noting the inequality C < A -f- tt, satisfied in general by moments of 
inertia, show that the semmnglo of the space cone cannot exceed 

tan- 1 { \/2. 

8. Make a rough estimate of the speed at which a twenty-five-cent piece 
must roll in order that its motion may be stable. 

9. A solid homogeneous cuboid of edges 2a, 2a, 4a can turn freely under 
no forces about its center, \\hich is fixed. It is set spinning with angular 
velocity u about a diagonal. Kind the semivertical angle of the cono 
described in space by the line through the center parallel to the longer 
edges, and show that the time taken by this line to move once round tho 
cone is 10jr/(w \/ll). 

10. A circular disk of mass m and radius a is made to roll without slipping 
in steady motion on a rough horizontal plane, its plane being vertical and 
its track a circle of radius b. It completes a circuit in time r. Reduce to a 
force at the center of tho disk and a couple, the force system (including weight 
and the reaction of the plane) which must act on the disk in order that 
this motion may take place. The components of the force and the couple 
are to be expressed in terms of a, &, r, M. 

11. A rigid body with an axis of symmetry can turn about its mass center. 
There acts on it a fnction.il couple, Xo, where o> is the angular velocity 
vector and X a positive constant Show that the axial component of angular 
velocity is reduced to half its original value in a time (C log. 2)/X, where C 
is the axial moment of inertia. If (/ exceeds the transverse moment of inertia 
A, show also that the semiangle of the body cone decreases steadily. 

12. An egg-shaped solid of revolution rolls in steady motion on a rough 
horizontal plane, the axis of figure being horizontal. Establish the relation 

(a 2 -f k*)ns - abn* + bg, 

where a is the radius of the greatest circular section, b the distance of the 
mass center from this section, A the radius of gyration about the axis of 
figure, and n, s the vertical and horizontal components of angular velocity. 

13. Prove that a top cannot move in steady precession with spin s and 
inclination to the vertical, unless 

C Y2 s 2 > 4Amga cos B. 

14. A solid cone of height b and semivertical angle a rolls in steady motion 
on a rough horizontal table, the line of contact rotating with angular velocity 



456 MECHANICS IN SPACE [Ex. XIV 

il. Show that the reaction of the table on the cone is equipollent to a single 
force which cuts the generator of contact at a distance 



'$ cos a H --- cot a 

from the vertex, whore k is the radius of gyration of the cone about a genera- 
tor. Deduce that the greatest possible value for 12 is 



as the cone would overturn ii S2 exceeded this value. 

15. A circular disk of radius a spins on a smooth table about a vertical 
diameter. Prove the motion is stable if the angular velocity exceeds 2 \/g/a* 

16. A circular disk of radius a rolls on a rough horizontal plane in steady 
motion. The speed of its center is q tt , and its plane is inclined to the vertical 
at a constant angle 0. Show that the radius r of the circle described by the 
center of the disk satisfies the equation 

4gr 2 Qqlr cot 6 qla cos 0; 
deduce that, when is small, 

3 '/.1 . i 

r = approximately. 

200 

17. A rigid body can turn freely about a smooth axis, for which its moment 
of inertia is 7. It is acted on by a couple of constant magnitude G, applied in 
such a way that, when the body has turned through an angle 0, the vector 
representing the couple makes an angle with tho axis. If the body is ini- 
tially at rest, find its angular velocity when it has turned through a right 
angle. 

If the axis is an axis of symmetry of the body, find the reaction exerted 
by the body on the axis when it has turned through an angle 0. 

18. A light axle L carries two gyroscopes; L is their common axis of sym- 
metry, and they can turn freely about it. L is so mounted that it can turn 
freely about a fixed point ^on it, halfway between the mass centers of the 
gyroscopes. Find a quadratic equation to determine the angular velocities 
of steady precession under the action of gravity, in terms of the following 
constants: 

m, w', the masses of the gyroscopes, 

C, C", their axial moments of inertia, 

Aj A', their transverse moments of inertia at their mass centers, 

8, s', their spins, 

2a, the distance between their centers, 

0, the inclination of L to the vertical. 

19. A rigid body turns about a fixed point under no forces. Show that, 
relative to the body, the extremity of the angular momentum vector h moves 



Ex. XIV] MOTION OF A RIGID BODY 457 

on the curve of intersection of the sphere 

* s + y z + s 2 = K 
and the cone 



the axes being principal axes of inertia, 

Sketch the curves, taking A > R > C and considering all possible values 
of 22'A 2 . 

20. A top is spinning about its axis which is vertical. It is at the same 
time sliding over a smooth horizontal plane with velocity q u . The vertex 
strikes a small smooth ridge on the plane, the direction of the motion being 
inclined at an angle a to the direction of the ridge. H the coefficient of 
restitution for the impact is e, find the angle at which the vertex rebounds 
from the ridge in terms of </ n , a, e, and the constants of the top. Find also 
the direction of motion of the mass center immediately after impnct 

21. A thin elliptical plate of semiaxos a, b (a > 6) can turn freely about its 
center, which is fixed; it is set in motion with an angular velocity n about an 
axis in its plane equally inclined to the axes of the ellipse. Show that the 
instantaneous axis will again be in the plane of the plate after a time 



where 



_ 

A "" 2 2 



22. A thin hemispherical bowl of mass m and radius a stands on a smooth 
horizontal table. A homontal impulsive force; of magnitude f* is applied 
along a tangent to the nm. Find the magnitude and direction of the veloc- 
ity instantaneously imparted to the point of the bowl in contact with the 
table. 

Show that, no matter how large P may be, the rim of the bowl will never 
come into contact with the table. 

23. A rigid body with an axis of symmetry k is mounted so that it can 
turn freely about its mass center, which is fixed. To a given point on the 
axis of symmetry there is applied a force FK, where K is a fixed unit vector 
and F a given function of the angle between k and K. Show that the 
system is conservative, and obtain equations analogous to (14.227) and 
(14.228). Show that k oscillates between two right circular cones having 
K for their common axis. 



CHAPTER XV 
LAGRANGE'S EQUATIONS 

16.1. INTRODUCTION TO LAGRANGE'S EQUATIONS 

The question must have occurred to many people: If science 
keeps on growing at its present rate, how are succeeding genera- 
tions of students to keep up with it? We may find a partial 
answer by looking back at what has happened during the past 
two hundred years or so. 

First, there has been the development of specialization a 
broad specialization into subjects (pure mathematics, applied 
mathematics, astronomy, physics, chemistry), followed by a 
narrower specialization into branches (differential geometry, 
hydrodynamics, spectroscopy, to mention a few). Each branch 
is now big enough to provide work for a lifetime. A still nar- 
rower specialization is not a pleasant prospect, for intensive 
work in a restricted range becomes in time uninteresting and 
sterile. 

But, side by side with the growth of specialization, we find an 
increasing tendency to use mathematical methods. Mathe- 
matics gives to science the power of abstraction and generalization, 
and a symbolism that says what it has to say with the greatest 
possible clarity and economy. The mathematician, penetrating 
deeply into the structure of theories, is often able to detect 
common features, not obvious on the surface. In this way, 
he breaks down the barriers between restricted fields and brings 
the specialists into contact with one another. Further, the 
mathematician can compress a mass of descriptive theory into 
a few differential equations and so greatly reduce the bulk of 
science. 

Long before science reached its modern state of complexity, 
Lagrange invented a uniform method of approach for all dynami- 
cal problems. This method has formed- the basis for nearly all 
work on the general theory of dynamics and is the foundation 

on which quantum mechanics is built. In the more elementary 

458 



SEC. 15.1] LAGRANGE'S EQUATIONS 459 

parts of mechanics, it has not yet supplanted the more direct 
and physical approach, because its rather abstract and general 
character has made it appear difficult. However, it seems 
probable that as time passes the method of Lagrange will work 
its way from the end to the beginning of textbooks on mechanics. 
The easier, but more cumbrous, methods are becoming a luxury 
for which we cannot afford the time. 

Instead of proceeding at once to Lagrange's equations in 
their full generality, we shall start with the case of a particle 
in a plane. Much of the difficulty of understanding the method 
may be overcome by a study of this comparatively simple case. 

Lagrange's equations for a particle in a plane. 

Consider a particle of mass m, moving in a plane. Let Oxy 
be rectangular Cartesian axes, and let X, Y be the components 
of the force acting on the particle. The usual equations of 
motion are 

(15.101) mx = X, my = Y. 

Now let qi, q% be any curvilinear coordinates (e.g., polar 
coordinates). It will be possible to express x and y in terms 
of <7i and q^ and so we may write 

(15.102) x = x(qi, (72), y = y(qi, <? 2 ). 
These relations hold for all values of the time t, and so 



The partial derivatives occurring here are functions of q\, q^ we 
can calculate them when the functions (15.102) are given. 

Looking at (15.103) in a formal way and forgetting that qi 
is actually the derivative of </i, we may regard them as equations 
expressing the two quantities JT, y as functions of the four quanti- 
ties qi, q 2 , qi, qz] we may express this by writing 

(15.104) * = f(q ly ? 2 , qi, &), y = flffai, 2, ffi, &) 
If we speak of the partial derivatives 

dx dx dx dx 

dqi dq* dqi dq 2 



460 MECHANICS IN SPACE [SEC. 15.1 

or the corresponding derivatives of y, we understand that they 
are calculated from (15.104), all the quantities q\, q^ ft, fa 
being treated as constants, except the one with respect to which 
we differentiate. 

But the functions in (15.104) are the same as those in (15.103), 
and so 



Furthermore, dx/dq\ is a function of </i, q 2 ; so, following the 
motion of the particle, 

d dx ' * x ' d 2 * . 



But, on the other hand, if we differentiate the first of (15.103) 
partially with respect to qi t we get 



/t r ^rtpr\ 

(15. 107) 



This is equal to the expression in (15.106). Hence, assembling 
this result with the other results obtained by using q% and y, 
we have 

^.^5. = ^ ^L^L^ d y. 

(15.108) dtd gi - dqi dtdqi'dqi 

d dx dx d dy dy 

dt dqz dq% dt dq% dq% 

The equations (15.105) and (15.108) are fundamental in the 
development of Lagrange's equations. 
The kinetic energy of the particle is 



(15.109) T = 

If we substitute for x and y from (15.103), we obtain a function 
of <?i, 72, q\, qz, 

(15.110) T = T(q l9 g 2 , ft, ft). 
Actually this function is of the form 

(15.111) T = iH! + 2/iftft 4 
where a, A, 6 are functions of q\ 9 q%. 



SEC. 15.11 LAGRANGE'S EQUATIONS 461 

Now, in (15.109), T is expressed as a function of x, y\ by 
(15.103), x, y are functions of qi, # 2 , qi, # 2 ; hence we obtain, 
using (15.105), 

n *> 119^ ~ 4- ^L **$ 

( * ^ dqi ~ dx dqi + dy dji 

- dx , . dy 
= mx~ h my - 



d dT . dx , .dy t . dx 
mx + m y ~ + mx 



Therefore, by (15.108) and (15.109), we havo 

(15.113) 

Subtracting and using the equations of motion (15.101), we get 



a?7 

r = mx - -- \-rny -- 
u 



There is, of course, a similar equation with <? 2 instead of qi. 

Any small virtual displacement* of the particle corresponds to 
increments dqi, dq 2 in the coordinates #1, q%. The corresponding 
increments in x, y are 

(15.115) te = | fe + g *, - g- ^ + | 2 5 9 , 

The work done in this displacement is 

(15.116) dW = Xdx + Ydy, 
or 

(15.117) dW = Qi dqi + Q 2 8q z , 
where 

(15.118) Q^X^ + Yg-, Q^xf- + Y^-. 

dqi dqi dq 2 dq 2 

Hence, we have the following result: The motion of a particle 
in a plane satisfies the differential equations 

* It should be emphasized that this virtual displacement is arbitrary; it 
is not to be confused with the displacement actually occurring in the motion. 
Tf we want to refer to the latter, we write dx, dy, dq\, dq*. 



462 MECHANICS IN SPACE [SEC. 15.1 

d dT dT _ n d dT dT - n 

dt djl ~ d ~ Ul > A *f 2 ~ 5^ ~ y " 

where q\, q% are any curvilinear coordinates, T is the kinetic energy 
(expressed as a function of q\, <? 2 , <h, #2) araZ Qi, Q 2 are obtained 
from the expression dW, as in (15.117), for the work done in an 
arbitrary small displacement. 

These are Lagrange's equations of motion. 

The curvilinear coordinates q\, q* are, of course, generalized 
coordinates, as discussed in Sec. 10.6; the quantities Qi, Q 2 are 
the generalized forces [cf. (10.708)]. 

It must be clearly understood that the Lagrangian method 
only provides the differential equations of motion; it does not 
solve them. It is true that the method does give some hints 
helpful for solution, but that is a matter into which we cannot 
go here. 

Example. Consider a particle of mass m moving in a plane, under an 
attractive force /xw/r 2 , directed to the origin of polar coordinates r, 9. If 
we take as generalized coordinates 

tf i = r, 92 = 0, 

and denote the generalized forces by ft, O, the equations of motion (15.119) 
read 



Now 

T = %m(r 2 + r 2 2 ), 
and so 

(15.121) = wir, = mrtf 2 , -^ = rar 2 0, - = 0. 
To find R and O, we have (for an arbitrary displacement 5r, 50) 

ft 5r -f O 5fl = dW == ~ 5r, 
and so 

(15.122) R - - ^~, 6 - 0. 

Substituting from (15.121) and (15.122) in (15.120), we get 

(15.123) mf mrd* = ^-r- -5- (mr 2 d) = 0. 

r z f dt 

These equations are the same as (5.104); the present method of obtaining 
them is simpler than the method used earlier. 



SEC. 15.2] LAGRANGE'S EQUATIONS 463 

16.2. LAGRANGE'S EQUATIONS FOR A GENERAL SYSTEM 

Lagrange's equations for a system with two degrees of freedom. 

We pass now from a particle moving in a plane to any system 
with two degrees of freedom, with generalized coordinates 
qit #2 (cf . Sec. 10.6). Let N be the number of particles forming the 
system, and let the Cartesian coordinates of a particle (of mass 
m t ) be Xi, 7/ t , Zi (i = 1, 2, N). Then Xi, y^ Zi are functions 
of gi, #2, and we may write 

(15.201) x % = Xi(qi, g 2 ), y t = 2/tfei, (72), 2t = ft(qi, #2). 

Here we have 3 AT equations like the two equations (15.102), and 
we obtain on differentiation 3JV equations like (15.103), 



(15.202) 

dZi 



As a matter of fact, the whole argument for a system with two 
degrees of freedom follows very closely the argument for a 
particle in a plane; the complication introduced by having 3N 
Cartesian coordinates, instead of only two, is not serious. Thus, 
if we use a symbol to stand for any one of the coordinates 
%i, 2/t, 2i, we obtain, exactly as in (15.105) and (15.108), the follow- 
ing equations: 

n - 9n ~v a _ a dt _ d( 

(15.203) - - 



^ ? q 

The kinetic energy of the system is 

N 

(15.205) T = i 2* m f fe 2 + t/? 

-i 

and this is expressible in the form 

(15.206) T = rfei, (?2, tfi, 



464 MECHANICS IN SPACE [SEC. 15.2 

As in the case of the single particle, this is a quadratic expression 

(15.207) T = %(aq\ + 2A44, + &#), 

where a, ^, b are functions of #1, q%. 
Then, by (15.203) and (15.205), 

ri5208^ ar - y /^^ , 

(15.208) - 2 + 



and so, by (15.204), 



The last term on the right is dT/dqi. Let X t , Y % , Z* be the com- 
ponents of force (external and internal) acting on the ith particle, 
so that 

(15.210) wiA = X t , my, = Y it mzi = 2 f . 
Then (15.209) may be written 



Now, if Qi, (h are the generalized forces, so that the work done 
in a general displacement is 

(15.212) BW = Qi Bqi + Q 2 tq*, 

it is clear from (10.707) that the expression on the right-hand 
side of (15.211) is precisely the generalized force Qi. 

Thus, associating with (15.211) the companion equation 
in 2 , we have Lagrange's equations of motion for a system with 
two degrees of freedom, 

(15213) ddT - dT - ddT -W_~ 

(15.216) QJ[-V>> J t ^ 3q 2 -V*' 

(SW = Qi S qi + Q 2 tqj. 



SEC. 15.2] LAGRANGE'S EQUATIONS 465 

The form of these equations is precisely the same as for a particle 
in a plane; no additional complexity has been added by consider- 
ing the general system with two degrees of freedom, of which a 
particle in a plane is, of course, a special case. 

When the system is conservative, with potential energy 
V(q\, #2), the generalized forces are connected with V by (10.712). 
Thus, (15.213) may be written 

. ([<W_dT = _ 37 ^^_i?! = _5?Z 
( ' dt dtfi dq l ~ dqi dt dq> 2 dq z dq z ' 

Example 1. Wo shall now find the equations of motion of a spherical 
pendulum. Let ra be the mass of the particle and a the radius of the sphere 
on which the particle is constrained to move. We take as generalized 
coordinates 



where is the angular distance from the highest point of the sphere and < 
the azimuthal angle. Then, 

T = $wa*(0 2 + sin 2 6 < 2 ), V = mga cos 0, 
and so 

AT 1 dT r)V 

- = ma*&, ma 2 sin cos <j> z , ~^r = mga sin 9, 
30 o0 o0 

g-, n..* g=0, g=0. 

Thus (15.214) give, as equations of motions of a spherical pendulum, 
ma?8 ma z sin cos < 2 = mga sin 0, 
5T (mo 2 sin 2 ^) =0. 

Example 2. Consider a uniform bar hanging by one end from a smooth 
horizontal rail. It can move only in the vortical plane through the rail and 
is under the influence of gravity and a horizontal force A' applied to its 
lowest point. Let us find the equations of motion. 

For generalized coordinates, we take 

QI = distance of point of suspension from some fixed point on rail, 

q z = inclination of bar to vertical. 
Then, 



n | [5 ( 



(a cos q 



where m = mass of bar, 
2a length of bar, 
k radius of gyration about mass center. 



466 MECHANICS IN SPACE [Sue. 15.2 

The above expression reduces to 

T = Mq\ + 2a cos q, qfa + (a 2 + *;)$}]. 
The generalized forces are given by 

Qi % -h Q 2 $?2 = X 8(qi + 2a sin g 2 ) + 
Thus 

Qi = X, <?2 = 2Xa cos 92 

and so the equations of motion are, by (15.213), 

m ~dt ^ l ^~ a C S <?2 ^ = ^' 

m -57 [a cos /? 2 #i -h (a 2 + & 2 )(?2] + ma sin q% qifa -2 Xa cos 72 wa0a sin q z . 

If the bar remains nearly vertical, so that g 2 is small, these equations 
simplify to the approximate form 



Lagrange's equations for a general system. 

Consider a system with n degrees of freedom and generalized 
coordinates qi, q^ * q n . The method of finding Lagrange's 
equations in this general case differs from the method given 
above only in a slightly greater complexity, due to the n degrees 
of freedom. We shall give here an argument complete in essen- 
tials but omitting details which can be supplied by the type of 
argument used earlier.* 

Let m ly x t , #t, z l (i = 1, 2, N) be the mass and coordi- 
nates of the ith particle. Then, for r 1, 2, - n, 

dTdXj dT dy, dT dz l \ 
^\dq r dy t dq r dZtdq r ) 



ddT 
dtddr 



* As in Sec. 10.6, non-holonomic systems will not be considered. 



/.. dx< .. dy t .. dz> 
V ' ^ ' ^ * d 



SEC. 15.3] LAGRANGE'S EQUATIONS 467 

This last equation may be written in the form 

d dT _ dT __ / Y dXi v dy t 7 dz*\ 

dtdq r dq r - ft^'dfr + 'Wr + *'Wr)' 

where X^ F t -, Z l are the components of force acting on the ith 
particle. 

Thus, by (10.707), we have Lagrange's equations of motion for 
a system with n degrees of freedom, 

! = <2" fr -1,2, ), 

where Q r are the generalized forces, defined by the condition that 
the work done in a general displacement is 

(15.216) 3W = J) Q r 8q r . 

r = 1 

// the system is conservative, 

(15.217) Q r = - | (r = 1, 2, n). 

Two features of Lagrange's equations should be emphasized. 
First, there is no unique set of generalized coordinates; however 
we choose them, the equations of motion always have the form 
(15.215). Secondly, since only working forces contribute to 8W, 
reactions of constraint are automatically eliminated.* 

16.3. APPLICATIONS 

Components of acceleration in spherical polar coordinates. 

Although the normal use of Lagrange's method is to obtain 
equations of motion, it may sometimes be used indirectly to give 
information not easy to obtain otherwise. Consider a particle 
moving in space. Let us take the spherical polar coordinates 
r, 6, as generalized coordinates qi, q%, # 3 . Then, if the particle 
is of unit mass, 

T = (r* + r*&* + r* sin 2 6 < 2 ). 

(To obtain this, we need only the components of velocity along 
the parametric lines.) If R, 0, * are the generalized forces, the 
equations of motion are, by (15.215), 
* Except where forces of friction do work. 



468 MECHANICS IN SPACE [SBC. 15.3 

r rd 2 r sin 2 < 2 = R, 

d 

-r (r 2 6) r 2 sin cos < 2 = 0, 

d / o o -x 

-E (r 2 sin 2 6 (p) = <. 

Let / r , /0, /^ be the components of acceleration along the para- 
metric lines. These are equal to the compo,nents of force in 
these directions. Hence, equating two different expressions 
for work done in an arbitrary displacement, we have 

dW = f r 5r + far 50 + far sin d<j> = R dr + 6 50 + * 5<, 
and so 

r = JK = r r0 2 r sin 2 d> 2 



(15.301) 



= 1 e = i ^ (r 2 0) - r sin cos < 

7* 7* nC 



= _- . 
r sin r sin eft 



These are the components of acceleration along the parametric 
lines of spherical polar coordinates. 

Normal frequencies of vibration of a system with two degrees 
of freedom. 

Let C be a position of equilibrium of a conservative system 
with two degrees of freedom. Let us choose generalized coordi- 
nates such that </i = <? 2 = at C. The kinetic energy is expres- 
sible in the form 



(15.302) T = 

where a, h, b are functions of </i, q z . Let us, however, consider 
only small oscillations about C, so that q\, q z , qi, (fa are small. 
Then the principal part of T has the form (15.302), where a, /i, b 
are constants, viz., the values of the coefficients for qi = q 2 = 0. 
Consider now the potential energy V. We may choose C 
as standard configuration, so that V = for q\ = #2 = 0. The 
expansion of V in a Taylor series reads 

(15.303) F-Zfc + t, 



SBC. 15.3] LAGRANGE'S EQUATIONS 469 

where the partial derivatives are evaluated for q\ = q% = 0. 
But, by the principle of virtual work (10.714), 



for #1 = #2 = 0. Hence the principal part of V is 
(15.304) V = \(Aq\ + 2H qi q 2 + Bq\), 



where A, H, B are constants. Thus, to our approximation, 
T and V are homogeneous quadratic forms with constant coeffi- 
cients, T being quadratic in the velocities and V in the coordinates. 
We have 

dT . , ,. dT dV A . u 

- offi + A*, g = 0, - Aft + Hq* 

dT ,. , ,. dT 7 A dF y/ , D 

= %x + 6^ 2 , = 0, = U qi + Bq 2 , 

and so Lagrange's equations read 



We seek a solution of the form 

</i = a. cos (tit + e), q 2 = /3 cos (n + e). 

When we substitute in (15.305) and eliminate a and /3, we obtain 
for n the determinantal equation 

(15.306) J*I- ^-^ =0. 

If Tii, n 2 are the roots of this equation, the normal periods (cf. 
Sec. 7.4) are 2ir/ni, 2ir/n^ and the normal frequencies arc ni/2ir, 



It is, of course, assumed that the equilibrium is stable. If 
it were not, we should discover the fact through the appearance 
of a zero or imaginary value for n. 

The top. 

Consider a top with fixed vertex 0. The system has three 
degrees of freedom. For two generalized coordinates, we take 
0, 0, the polar angles of the axis of the top, 6 = being directed 



470 MECHANICS IN SPACE [SEC. 15.3 

vertically upward. For the third coordinate, we take the angle 
^ between two planes, one fixed in the top and passing through 
its axis, and the other containing the vertical through and the 
axis of the top. Then the angular velocity has components 0, 
sin 6 <, at right angles to one another and to the axis of the top, 
and a component 4> + cos 6 < along the axis. Thus the kinetic 
energy is 



(15.307) T = %A(6* + sin 2 < 2 ) + |C(^ + cos < 2 , 

where A and C are the transverse and axial moments of inertia 
at the vertex. The potential energy is 

(15.308) V = mga cos 0, 

where a is the distance of the mass center from the vertex. 
Lagrange's equations then read 

A6 - A sin cos < 2 + C sin <(^ + cos <) 

= mga sin 0, 
d 



~ [A sin 2 < + C cos 0(^ + cos <)] = 0, 
+ cos ] = 0. 



(15.309) 



The last two equations give at once the first integrals 

/i e oim ( A sin2 e * + C cos *<* + cos **) = 
(15.310) | ^ + CQS ^ _ ft 

where a and are constants. (These are actually integrals of 
angular momentum.) When we substitute in the first of (15.309), 
we get a differential equation for 



(15.311) 9 + a ~ * = sin 0. 

If we multiply this equation by 0, integrate once, and put 
cos = #, we get (14.226). The detailed theory of the motion 
then proceeds as in Sec. 14.2. 

Lagrange's equations for impulsive forces. 

When impulsive forces act, there are instantaneous changes 
in velocity, without instantaneous changes in position. In 
terms of generalized coordinates q r , there are instantaneous 



SEC. 15.3] LAGRANGE'S EQUATIONS 471 

changes in q r , but not in q r . As usual, we approach impulsive 
forces by a limiting process, in which the forces tend to infinity 
and the interval during which they act tends to zero. We 
multiply (15.215) by dt and integrate over the interval (o, ti). 
When ti Jo, the second term on the left disappears, and we 
have Lagrange's equations for impulsive forces, 

(15.312) A ff = &' (r-1,2, -n); 

here A denotes a sudden increment and Q r are the generalized 
impulsive forces [cf. (8.112)] 

(15.313) Q r = lim f 11 Q r dt. 

*i-Xo J** 

These may be calculated from a formula analogous to (15.216), 

(15.314) 8W = 



where 8W is the work which would be done in a general dis- 
placement by the impulsive forces if they were ordinary forces. 

The Lagrangian method is particularly useful for systems of 
linked rods, because the impulsive reactions are automatically 
eliminated. Thus, to take an example, consider the problem 
worked in Sec. 8.3 (Figs. 99a and 996). As generalized coordi- 
nates we take x, y, the coordinates of the joint, and 0i, 2 , the 
inclinations of the rods to their initial line. Then, for the given 
position (6 i = 02 = 0), 

T = $m[x* + (y- orf,) a + k*6\ + x* + (y + a0 2 ) 2 + k*t}], 

where k is the radius of gyration of a rod about its center. Now 
if X, F, 61, 6 2 arc the generalized impulsive forces, we have 

2 5x + Y by + 61 50i + 02 502 = P(8y + 2a 50 2 ). 
Thus 

X = 0, f = P, 81 = 0, 82 - 

Lagrange's equations give 

2mx = 0, 

m(y - a6i) + m(y + a0 2 ) = ?, 
-ma(y - a^i) + mk 2 6i = 0, ^ 
ma(y + a6*) + mk*6 z = 2aP. 



472 MECHANICS IN SPACE [SEC. 15.4 

Hence we obtain, with /c 2 = a 2 /3, 

P P P 

x = 0, y = --> ^ =-.-, 2 = _.. 

' m ma ma 

16.4. SUMMARY OF LAGRANGE'S EQUATIONS 

I. Finite forces. 

For system with kinetic, energy T, expressed as function of 

(15.401) 4^ ~ %r = On (r = 1, 2, n). 

V 7 didQ'r d^ r ' V ' ' 7 

n 

(15.402) dW = ^ Q r 5g r . 

(15.403) Qr = ~ T > for conservative system. 

II. Impulsive forces. 

(15.404) A |? = Q r , (r = 1, 2, n). 

C/O'j' 
n 

(15.405) 5^ = ] Qr fy r . 



EXERCISES XV 

(To be done by Lagrange's equations) 

1. Find the equation of motion of a simple pendulum, taking in turn the 
following generalized coordinates : 

(i) the angular displacement, 
(ii) the horizontal displacement, 
(iii) the vertical displacement. 

2. Find the equation of motion of a sphere rolling down a rough inclined 
plane. 

3. Find the equations of motion of a spherical pendulum, taking as 
generalized coordinates the horizontal Cartesian coordinates of the bob. 
Reduce the equations to their principal parts for oscillations near the equi- 
librium position. 

4. Four flywheels with moments of inertia /j, /2, Is, /4 are connected by 
light gearing so that their angular velocities are in fixed ratios n\: nz'.n^n^. 
Driving torques #1, #2, N*, Nt are applied to the flywheels. Find their 
angular accelerations. 

5. Show that if a generalized coordinate (q\) does not appear explicitly 
in either T or V f then dT/dqi is constant throughout the motion. 



Ex. XV] LAGRANGE'S EQUATIONS 473 

A rod hangs by a universal joint from its upper end. For oscillations 
under gravity, use the above result and the equation of energy to find a 
differential equation of the first order for 0, the inclination of the rod to the 
vertical, 

6. A pendulum consists of two equal bars AB, BC, smoothly jointed at B 
and suspended from A. The mass of each bar is m, and its length is 2a. 
Find the normal periods for small oscillations in a vertical plane under 
gravity, in the form 



where X is a numerical constant. 

7. The pendulum described in Kxercise 6 hangs at rest. A horizontal 
impulse P is applied at its lowest point. Find the angular velocities 
imparted to the bars. 

8. The ends of a heavy uniform bar of mass 120 Ib. are supported by 
springs of equal strength, the bar being horizontal. The strength of the 
springs is such that a weight W of 100 Ib , placed gently at the middle point 
of the bar, causes it to descend 1 in. Find, to two significant figures, the 
normal frequencies of small vibrations of the bar (without the weight W), 
considering only vibrations in which the springs move vertically 

9. On a sphere, and < are polar angles. A particle describes a small 
circle = constant with constant speed </ . Find the generalized forces O, 4 
consistent with this motion. 

10. (\>. ,.vicr a dynamical system with kinetic and potential energies 



where / is a given function. By choosing suitable new coordinates q{, g, 
reduce the problem of determining the motion to the evaluation of an 
integral involving the function /. Determine q i} q z as functions of t if 
/(*) - x\ 

11. A carriage has four wheels, each of which is a uniform disk of mass m. 
The mass of the carriage without the wheels is M. The carriage rolls 
without slipping down a plane slope inclined to the horizontal at an angle 
a, the floor of the carriage remaining parallel to the slope. A perfectly 
rough spherical ball of mass m' rolls on the floor of the carriage along a line 
parallel to a line of greatest slope. Show that the acceleration of the carriage 
down the plane is 

7M + 28m + 2m' 

TFT- 42wT+2m' ' Sm "' 

and find the acceleration of the ball. 

12. A rhombus of equal rods, smoothly jointed, lies on a plane in the form 
of a square. An impulse is applied to one corner, along the diagonal through 
that corner. Find the angular velocities imparted to the rods, in terms of 
the impulse (/*), the mass (m) of a rod, and the length (2a) of a rod. 



474 MECHANICS IN SPACE [Ex. XV 

IS. A smooth circular wire carries a bead. The wire is suspended from a 
point on it. Find the normal periods of small vibrations under gravity 
when the wire swings in its own plane and the bead slides on the wire. Show 
that, when the bead is fixed to the wire at its position of equilibrium when 
free to slide, the period coincides with one of these two normal periods. 

14. Using the fact that T is a homogeneous quadratic expression in the 
generalized velocities g r , show that the integral of energy T -f V = constant 
may be proved as a mathematical deduction from La grange's equations. 
Note that, if / is a homogeneous function of degree m in x\,x% x n , then 



r=s 1 

16. A dynamical system has kinetic energy 

T - 
and potential energy 

V - 
Additional generalized forces 



are applied. All the coefficients a, h, 6, A, H, B, and the k's are constants. 
Show that the energy sum T + V decreases steadily during any motion, 
provided 

kn > 0, ifciifc,, > (fci, + /b 21 ) 2 . 

16. A gyroscope is mounted in a light Cardan's suspension (Fig. 144). 
Take Eulerian angles simply related to the suspension, and find the equations 
of motion of the system under the action of a couple G applied to the outer 
ring, G being in the line of the outer bearings. 

17. A system is said to have "moving constraints" when tho configura- 
tion of the system is determined by the values of generalized coordinates 
#i #2, * <?n and the value of the time t. Show that, for such a system, 
Lagrange's equations hold in the same form as when there are no moving 
constraints, but that the kinetic energy is no longer a homogeneous quadratic 
expression in ,, 2 , #. 

Apply this result to find the equation of motion of a heavy bead on a 
smooth circular wire, the wire being made to rotate about the vertical diam- 
eter with constant angular velocity. 



CHAPTER XVI 
THE SPECIAL THEORY OF RELATIVITY 

16.1. SOME FUNDAMENTAL CONCEPTS 

The hardest part of a subject is the beginning. Once a certain 
stage is passed, we gain confidence and feel that, if need be, we 
could carry on by ourselves. The process of learning is very 
much the same whether in swimming or in mechanics an initial 
feeling of insecurity is followed by a feeling of power. 

The simple things that we learn first are the hardest to change 
later. Whether they are muscular actions or mental concepts, 
they are used again and again until they become part of us. Our 
bodies or minds have learned to follow a pattern, which can be 
broken only by a conscious effort. 

Breaking up the Newtonian pattern. 

We are now faced with the task of breaking up the pattern of 
Newtonian mechanics, to make way for the new pattern of 
relativity, which we owe to Einstein. This would be com- 
paratively easy if it were merely a question of making changes 
in the later and more elaborate parts of the subject. But that 
is not the case. The change is to be made right down in the 
foundations in our concept of time. 

To show how fundamental the change is, we shall describe an 
imaginary experiment, putting into opposition the predictions 
that would be made by a follower of Newton on the one hand and 
a follower of Einstein on the other. 

Two clocks stand side by side at a place P. They are of 
the very finest construction and identical with one another. 
Their readings are the same, and they continue to run in perfect 
unison as long as they stand side by side at P. One clock is 
left at P; the other is put in an airplane and flown with great 
speed on a long flight, being finally brought back to P and set 
up beside the clock that has stood there unmoved. 

Will there then be any difference between the readings of the 
two clocks? 

475 



476 MECHANICS IN SPACE [SEC. 16.1 

The practical physicist will, before answering, make inquiries 
as to the way in which the clock was treated on the flight 
whether it was knocked about, whether it was subjected to 
extremes of heat and cold, and so on. Let us suppose that the 
greatest care has been taken, so that effects due to these acci- 
dental causes may be ruled out of consideration. Then the 
answers are as follows: 

Newtonian theory : The clocks will show the same reading. 

Relativity theory: The readings will not be the same. The 
clock that has been on the flight will be slow in comparison with 
the clock that has stayed at home. 

The follower of Newton reasons along these lines: A perfect 
clock registers the time. A flight in an airplane does not alter 
this fact, provided that proper precautions are taken. Since 
after the flight each clock registers the time, they must agree. 

We cannot yet give the reasoning of the relativist; that 
will come later in the chapter. For the present, we must be 
satisfied with the words with which the relativist would begin 
his attack on the argument of the Newtonian: There is no such 
thing as the time, in any absolute sense. 

It would be impossible to decide between the two predictions 
by carrying out the experiment we have described. The rela- 
tivist would predict a difference between the two readings far 
too small to detect. The airplane would have to fly with a speed 
comparable with that of light before the effect would be notice- 
able. But it is the principle that is important. Other experi- 
ments can be carried out in which the predictions of the two 
theories are different and the difference is large enough to 
measure; in every case the relativistic prediction proves correct. 
There can be no doubt that the theory of relativity gives us a 
mathematical model closer to nature than the Newtonian model. 
We must therefore pay attention to the words: There is no such 
thing as the time, in any absolute sense. Once that point is 
conceded, the basis of the Newtonian pattern is broken, and the 
way is open for relativity. 

The ingredients of relativity. 

The theory of relativity is divided into two parts: 
(i) the special theory; 
(ii) the general theory. 



SEC. 16.1] THE SPECIAL THEORY OF RELATIVITY 477 

The special theory deals with phenomena in which gravitational 
attraction plays no part, while the general theory might be called 
"Einstein's theory of gravitation/* In this book, we shall be 
concerned solely with the special theory. 

At this stage the reader should glance over Chap. I to concen- 
trate his attention again on fundamental matters. Part, but 
not all, of the contents of that chapter will pass over into the 
theory of relativity, and we must understand clearly what 
passes over and what does not. Let us therefore start again with 
a blank sheet and put in, one by one, the ingredients of the 
theory of relativity. 

First we introduce a particle, understood in the same sense 
as before. Next we introduce & frame of reference and an observer 
in it. The observer has a measuring rod with which he can 
measure the distances between the particles which form his 
frame of reference. If the distances between these particles 
remain constant, the observer declares that his frame of reference 
is a rigid body. 

Now we provide the observer with a clock. As in Chap. I, 
this is an apparatus in which the same process is repeated over 
and over again, the repetitions defining equal intervals of time. 
The actual mechanism of the clock does not matter. We may 
think of it as an ordinary watch, driven by a spring and controlled 
by an escapement. 

As we have indicated above, the transporting of a clock is an 
operation which may lead to curious consequences. We shall 
therefore not expect the observer to carry his clock about but 
shall provide him with a great number of clocks, all of identical 
construction. These will be distributed throughout his frame 
of reference and kept fixed in it. 

We must not overlook the fact that the synchronization of 
these clocks raises an important and difficult question. If there 
is no synchronization, the observer will note some strange things 
as he walks among his clocks. For example, he may start at 
2:15 (by the local clock), walk a mile, and find that the time is 
2:10 (by the local clock). Under such circumstances, in ordinary 
life, one would put a clock in his pocket and walk around, setting 
each local clock as he passed to agree with the clock in his pocket. 

But if our observer does this he finds the following strange 
result. The clocks which he synchronizes in walking out from 



478 MECHANICS IN SPACE [SEC. 16.1 

his base no longer agree with the clock in his pocket when he is 
walking back. 

This is the same phenomenon as that described earlier in the 
case of the clock and the airplane, and the reason for it will be 
made clear later. The effects are so small as to be negligible 
in ordinary life, but our observer is expected to be mathe- 
matically accurate. 

This description of the difficulties of synchronization may 
explain why the theory of relativity has had for the popular 
mind much the same appeal as Alice in Wonderland. Familiar 
ideas are turned upside down. Why does the observer not simply 
set all the clocks to show the correct time? The answer is: There 
is no such thing as the correct time. 

We recall that, in Chap. I, we introduced the idea of an event 
something happening suddenly at a point. We carry this idea 
over into relativity, where we shall make extensive use of it. 
Even though his clocks are not yet synchronized, the observer 
is prepared to describe any event by assigning four coordinates 
to it. Of these coordinates, three are spatial (x, y, z), and the 
fourth (t) is given by the local clock, i.e., the clock situated at 
the point where the event occurs. 

Galilean frames of reference. 

We have already seen in Newtonian mechanics the importance 
of making a proper choice of frame of reference. The laws of 
Newtonian mechanics take their simplest form only in certain 
special frames, which we called Newtonian. Similarly, in rela- 
tivity there are frames of reference which are particularly 
convenient to use. These are called Galilean frames of reference. * 
They correspond in nature to rigid bodies situated in remote 
space, far from attracting matter, and without rotation rela- 
tive to the stars as a whole. 

We shall now make the following hypothesis regarding a 
Galilean frame of reference: 

I. A Galilean frame of reference is a rigid body, isotropic with 
respect to mechanical and optical experiments. 

* This is the usual name, and not a very good one, for Galileo lived before 
Newton and of course had no idea of the theory of relativity. "Einstein 
frame of reference" would be a better name. 



SBC. 16.1] THE SPECIAL THEORY OF RELATIVITY 479 

To explain this, we note that "isotropic" means "the same 
in all directions." The neighborhood of the earth is not iso- 
tropic. If we drop a stone, it falls in a definite direction and 
the earth's rotation defines a direction which we can detect by 
means of a gyrocompass. However, in applying the theory of 
relativity, we may often regard the earth as a Galilean frame, 
for its gravitational attraction may be small compared with other 
forces involved and its rotation may be of no importance. 

The assumption that a rigid body in remote space is iso- 
tropic is at least plausible. To assert that it was not isotropic 
would at once raise the question: Why should any one direction 
be privileged above another? 

The hypothesis refers to mechanical and optical experiments. 
We must provide the observer with apparatus to perform these. 
We shall therefore give him mechanisms by which he can exert 
forces, and lamps and mirrors by which he can send out flashes 
of light and reflect them. 

In Newtonian mechanics, we had no occasion to refer to light. 
Optics appeared to be a separate subject. In relativity, on the 
other hand, we have to discuss optics and mechanics together. 

The synchronization of clocks. 

Space does not permit us to attempt an axiomatic treatment of 
the theory of relativity. To reach the most interesting deduc- 
tions quickly, we shall outline some steps in the development 
without proof. 

Thus, we shall only sketch the method of synchronization of 
clocks in a Galilean frame of reference. The synchronization 
is done by means of light signals. Taking the clock at the origin 
as master clock, the observer sends out flashes of light to the 
other clocks, from which they are reflected by mirrors back to 0. 
Let ti and 2 be the times (as given by the clock at 0) at which a 
flash leaves and returns to it after reflection at a point A. 
In ordinary life, we should reason in this way : If v is the velocity 
of light and r the distance OA, the light would take a time r/v 
to go and a time r/v to return. Thus t z t\ = 2r/v, and the 
time of arrival at A is 

t = ti + r/v = ti + i( 2 - i) = i(i + it). 
But we cannot use this argument, because velocity is a derived 
concept, depending on the measurement of both distance and 



480 MECHANICS IN SPACE [Sac. 16.2 

time. We should be arguing in a circle if we used velocity to 
define time. We shall merely adopt as definition of synchroniza- 
tion that the clock at A is synchronized when it is set to read 
s(ti + 2 ) at the instant when the flash strikes it. By this rule, 
all the clocks may be synchronized with the clock at 0. 

We ask the reader to accept the fact that, in consequence of 
the assumption of isotropy, this synchronization is satisfactory. 
That is, a repetition of the process, with another clock as master- 
clock, will find all clocks reading just what they ought to read, 
so that no change in the settings is necessary. This means that 
there is no confusion such as we predicted earlier, in the case 
where the synchronization was attempted by carrying a clock 
about. 

The observer now has a serviceable time system. He can 
measure velocities, and in particular the velocity of light. By 
virtue of the assumed isotropy, this proves to be a constant, 
the same for all directions. 

Although we have met some new ideas in connection with 
synchronization, there is nothing new or strange about the final 
picture of a Galilean frame of reference and the time system we 
have set up in it. It differs in no essential way from the concept 
we have used in Newtonian mechanics. We do not encounter 
the real peculiar! tios of relativity until we consider two Galilean 
frames of reference and the relations between them. 

16.2. THE LORENTZ TRANSFORMATION 
The principle of equivalence. 

Let us suppose that we can shoot a rocket right out of the 
solar system. In this rocket we place an observer. When the 
io'..ket has passed far beyond the solar system, it forms a Galilean 
frame of reference. The observer has lost. the sense of motion 
he had when rushing past the planets. He sees around him 
nothing but stars, and they are so far away that they appear 
fixed. 

A second identical rocket is shot out with a greater speed and 
on such a track that it overtakes the first. In it there is also an 
observer. Now we have two Galilean frames of reference. ' 

Imagine that the two observers leave their rockets and travel 
independently in space. One of them comes upon one of the 
rockets. How is he to tell whether it is the rocket he occupied 



SBC. 16.2] THE SPECIAL THEORY OF RELATIVITY 481 

before or the other one? To answer this question, he is allowed 
to perform any mechanical or optical experiments he chooses. 

The same question in a different form occurred to the physicist 
Michelson in 1881. What he asked might be put thus: Is it 
possible to tell the season of the year (i.e., the position of the 
earth in its orbit round the sun) by means of optical experiments 
performed on a clouded earth? The earth at the two seasons 
corresponds to the two rockets (Galilean frames of reference). 
It was fully expected that the season could be determined in this 
way, for it was then believed that light was propagated in an 
"ether," and the difference between the velocities of the earth 
through the ether at the two seasons should be a measurable 
quantity. 

The question was put to experimental test by Michelson 
and later by Michelson and Morley in 1887.* The expected 
result was not obtained. As far as this experiment was con- 
cerned, the two seasons (Galilean frames of reference) were 
indistinguishable. 

Generalizing from the negative result of the Michelson- 
Morley experiment, we make ohe f Blowing sweeping hypothesis: 

II. PRINCIPLE OF KQUIVA.LEN ,E. Two Galilean frames of 
reference are completely equivalent for ALL physical experiments. 

This gives the answer to the question raised earlier. The 
observer is not able to tell which rocket he has found. No 
experiment he can perform will tell him which it is they are 
indistinguishable, like identical twins. 

To the hypothesis already made we add another: 

III. Any two Galilean frames of reference have, relative to one 
another, a uniform velocity of translation. The relative velocity 
is less than the velocity of light. 

We may recall that, in Newtonian mechanics, two Newtonian 
frames of reference are similarly related, but in that case there is 
no restriction on the relative velocity. The assumption that the 
velocity of light is a limit which cannot be exceeded is something 
essentially new. 

To sum up, we have made three hypotheses in all. The first 
deals with a single Galilean frame of reference; the last two con- 
cern the relations between two Galilean frames of reference. 

* For an account of the Michelson-Morley experiment, see L. Silberstein, 
The Theory of Relativity (Macmillan Company, Ltd., London, 1924), p. 71. 



482 MECHANICS IN SPACE [SBC. 16.2 

The Lorentz transformation. 

Let S and S' be two Galilean frames of reference. (We 
may without confusion also use these letters for observers in 
the two frames.) Consider any event, observed by both observ- 
ers. To this event, S attaches coordinates (x, y, z, t), and S' 
attaches coordinates (z', y', 2', t'}. In doing this, each observer 
uses his own measuring rod and clocks. The. event determines 
the coordinates, and conversely the coordinates determine the 
event. Thus, considering all possible events, four numbers 
(x, y, z, t) determine an event, and that in turn determines the 
four numbers (x 1 , y\ z', t f ). The last four numbers are therefore 
functions of the first four, and we express this by writing 

(16.201) x' = f(x, y, z, 0, y' = 9(x, y, z, t), 

z' = h(x, y, z, 0, ? = l(x, y, z, t). 

Such relations, connecting the coordinates of two Galilean observ- 
ers, constitute a Lorentz transformation. 

We have now to investigate the forms of these functions. We 
shall not, however, suppose that the axes Oxyz and O'x'y'z' 
are given arbitrary directions in the respective frames. We shall 
suppose them so chosen that Ox and O'x' lie on a common line 
when viewed by either observer, this line being parallel to the 
relative velocity of either frame with respect to the other. We 
shall .consider the Lorentz transformation only for events occur- 
ring on this common line. Thus y = z = y f = z' = 0, and the 
transformation is of the form 

(16.202) x' - f(x, 0, t' = l(x, t). 

We must carefully avoid the idea that there is any " absolute " 
frame from which S and S f may be viewed. We must look at 
things either as they appear to S or as they appear to S'. First, 
S sees the particles of his own frame. They are fixed as far 
as he is concerned, and through them there pass the particles of 
the frame S'. These particles all move parallel to Ox with a 
constant speed 7, the speed of S' relative to S. Similarly, to 
S' the particles of his frame appear fixed, and the particles of S 
pass with a speed V, directed in the negative sense of O'x'. 
To do justice to both observers, it is best to draw two diagrams, 
as in Figs. 159a and 6. In Fig. 159a we take the view of S and 



SEC. 16.2J THE SPECIAL THEORY OF RELATIVITY 



483 



regard Oxyz as stationary; in Fig. 1596 we take the view of S' 
and regard O'x'y'z' as stationary. 

The two observers now fix their attention on a flash of light 
traveling along the common line Ox, O'x 1 . It will add to the 
complexity of our work if we assume that the units of space and 
time used by S and S' are completely independent. We shall 
therefore suppose that they have been supplied with measuring 
rods and clocks from a common stock. Then, by virtue of the 
principle of equivalence, the speed of light* has a common value 
in the two frames. This common value we shall denote by c. 



,/y 

If T f 



Z Z r Z Z' 

(a) (&) 

Fio. 169. (a) The frame S' moving relative to the frame S. (&) The frame <S 
moving relative to the frame S'. 

As S observes the flash, he records the time t at which (by 
his local clock) the flash reaches the position x. Since the speed 
of light is c, x is a function of t satisfying 

(*)-- 

But similarly, for the observations of $', 



dt' 

Thus, for the sequence of events given by the passage of the 
flash, we have the two equations 

dt* - dx*/c* - 0, dt f * - dx' z /c* = 0. 

Now a motion satisfying either of these equations represents the 
passage of a flash of light and therefore must satisfy the other 

* 3.00 X 10 10 cm. seer 1 or 186,000 mile seer 1 



484 MECHANICS IN SPACE [SEC. 16.2 

equation. Thus, if one of the equations is satisfied, so is the 
other, and therefore we have the identity 

(16.203) dt' 2 - dx' 2 /c 2 = k (dt 2 - dx 2 /c 2 ), 

where k is some unknown factor. But by the principle of equiv- 
alence we must also have 

(16.204) dt 2 - dx 2 /c 2 SEE k (dt' 2 - dx' 2 /c*). 

Comparing these two identities, we see that k 2 = 1, and so 
k = -f-1 or 1. To see which value to take, we follow the 
particle 0', fixed in 8'. Then dx' = 0, and so, by (16.203), the 
history of 0' satisfies 



But, by hypothesis III, (dx/dt) 2 < c 2 ; hence, k = +1. Accord- 
ingly, 

(16.205) dt' 2 - dx' 2 /c 2 = dt 2 - dx 2 /c 2 . 

The Lorentz transformation must be such that this identity holds. 
We shall assume that the transformation is linear, and that the 
zeros of time are chosen so that t = t' = when 0' is passing 
through 0. Thus we write, in place of (16.202), 

(16.206) x' = ax + pt, t' = a'x + P't, 

where a, p, a', 0' are constants so connected that the identity 
(16.205) is satisfied. We have 

!dx' = a dx + dt, dt' = a' dx + p' dt, 
dt' 2 - dx' 2 /c 2 = (a' dx + p' dt) 2 - (a dx + p df) 2 /c 2 
= dt 2 - dx 2 /c 2 , 

and so, equating the coefficients of dx 2 , dt 2 , and dx dt, 

(16.208) a 2 - cV 2 = 1, P 2 - c 2 P' 2 = -c 2 , op - c V/3' = 0. 
Let us define <t> } 0' by the equations 

(16.209) sinh = ca', sinh 0' = p/c. 
Then, by the first two equations in (16.208), we have 

a = cosh 0, p' = cosh <#>', 



SEC. 16.2] THE SPECIAL THEORY OF RELATIVITY 485 

and the last of (16.208) gives 

sinh (<' - 0) = 0. 
Thus <j>' = <, and the transformation (16.206) is 

(16 210} \ x r = x cosh 4> + ct sinh <, 

\ ct r = x sinh ^ + ct cosh 0. 

It is easy to verify directly that this transformation satisfies 
(16.205) for any constant <#>. 

To identify <, we take the viewpoint of S and follow the 
particle 0'. For 0' we have x f = 0, dx/dt = 7, and so differen- 
tiation of the first of (16.210) gives 

(16.211) tanh* = -V/c; 

hence, 



(16.212) cosh = 



So we have the Lorentz transformation 

(16.213) x' = y(x - 70, *' = 7 [t - 

1 

7 

Solving for j, , we get 

(16.214) x = 7(z' " 

If we now take the viewpoint of S r and follow 0, we have x = 0, 
dx r /dt' = 7', where 7' is the speed of $ relative to S'. But 
when we put x in the first of (16.214) and differentiate, we 
obtain dx' /dt' = 7. Hence V 7, as indeed we might have 
anticipated from the principle of equivalence. 

Now we have the explanation why the theory of relativity did 
not force itself on the attention of mankind long ago. Apart 
from the high velocities of electrons, which were not observed 
until comparatively recent times, physicists and astronomers 
have had to deal only with relative velocities very small indeed 
compared with the velocity of light. If V/c is small, then 7 is very 



486 MECHANICS IN SPACE [Sflc. 16.2 

nearly unity; as F/c 0, the Lorentz transformation (16.213) 
tends to 

(16.215) x' = x - F, *' = t, 
as in Newtonian mechanics (cf. Sec. 5.3). 

Immediate consequences of the Lorentz transformation. 

To a person accustomed to thinking in the Newtonian way, 
some of the predictions of the theory of relativity are startling. 
Outstanding among these are the contraction of a moving body 
and the slowing down of a moving clock. These apparently 
curious facts are consequences of the Lorentz transformation 
(16.213). 

First, let us consider a measuring rod which 5' lays down 
along his axis O'x'. To him it is a fixed measuring rod. If A, B 
are its ends, the history of A is a sequence of events for which 
x' = (X')AJ a constant, and the history of B is a sequence of 
events for which x' = (x f ) B , also a constant; the length of the 
rod is 

(16.216) L' = (x') B - (x') A . 

Viewed by S, the rod is not fixed. An instantaneous picture, 
taken by S at time t, shows A at (x) A , say, and B at (x) B . If 
asked what is the apparent length of the rod, S naturally says 
that it is 

(16.217) L = (x) B - (x) A . 
Now, by the first equation of (16.213), 

n2itt /(*'). = ?[(*). - Fl, 

(16.218) | (x% = 7[(x)A _ y t]i 

and subtraction gives, in view of (16.216) and (16.217), 

L' = tL, 
or 



(16.219; L = Z//T = L' VI - F 2 /c 2 < L'. 

Thus L is less than Z/; the rod appears to S to be contracted in 

the ratio \/l - V 2 /c 2 : 1. 

We might expect that, if S f viewed a rod fixed in $, he would 
see an expansion instead of a contraction. But if we carry out 



SEC. 16.2] THE SPECIAL THEORY OF RELATIVITY 487 

the calculation, now using (16.214) instead of (16.213), we find 
the same contraction again. Each observer considers that the 
measuring rod of the other is contracted. 

Now let us consider a clock carried along in S r . Let (')A, 
(t') B be two readings of the clock. These are two events, both 
with the same #', and with t' = (t') A , t r = (t') B , respectively. 
Viewed by S, the clock is moving; let the times of the two events 
be (t) Al (t)n, respectively, as measured in his time system. From 
the second equation of (16.214), we have 



r v\ 

A = 7 [')-. H-^-J- 



By subtraction, 

(0. - M> = T[')B 

or 

(16.221) T = yT> 



whore T is the time interval recorded by S and T' the time 
interval recorded by 8'. Since T' < T, the clock carried by 
S' appears to 8 to be running slow. Just as in the case of 
contraction of length, this result works both ways. Each 
observer considers the clock of the other to be running slow. 

Space -time. 

It does not seem possible at first sight to do justice simul- 
taneously to each of two Galilean observers, for it appears neces- 
sary to take the point of view of either the one or the other. 
This difficulty is overcome by using a space-time diagram. 

There is nothing peculiarly relativistic about a space-time 
diagram. We have used the idea in Newtonian mechanics, 
as in Fig. 78, when we plotted the position of a damped harmonic 
oscillator against the time. But it is in relativity that we get 
full advantage from this idea. 

Consider first one Galilean observer 8. Draw oblique Car- 
tesian axes on a sheet of paper, and label them Qx, fit (Fig. 160). 
(We use Q instead of for origin, to avoid confusion with the 
origin of the observer 's axes.) Any event which happens on the 



488 



MECHANICS IN SPACE 



[SEC. 16.2 



axis of Ox in the observer's frame will have attached to it values 
of x and t. It can then be represented by a point in Fig. 160, 
which is our space-time diagram. The history of a particle moving 
along Ox will appear as a curve C in the space-time diagram. 
If it moves with uniform velocity, dx/dt is a constant, and the 
curve becomes a straight line C\. 

Consider now a second Galilean observer S'. Instead of 
making a new space-time diagram for him, we superimpose his 
diagram on that of S. But we use new oblique axes $lx't f 
(Fig. 161), so related to ttxt that the geometrical law of transfor- 
mation of coordinates in the plane is precisely the Lorentz 
transformation (16.213). Now any event E has, of course, two 




Fia. 160. Histories of particles 
in the space-time diagram. 




Fio. 161.- An event 
and the space-time axes of 
two Galilean observers. 



pairs of labels (x, ), (#', t')\ but since these are connected by the 
Lorentz transformation, E appears as a single point in the space- 
time diagram. 

In fact, the space-time diagram gives us a representation of 
events independent of the frame of reference. It is the same 
situation as we have in geometry. The sides of a polygon drawn 
on a plane have equations which depend on the choice of axes. 
The polygon itself is something absolute. 

We must not, however, rush to the conclusion that the ordinary 
methods of geometry can be carried over completely into the 
plane of the space-time diagram. For example, in ordinary 
geometry we are accustomed to speak of the distance between 
two points as something independent of the axes used. If we 
have two sets of rectangular axes Oxy, Ox'y f in a plane, then the 
square of the distance between adjacent points is 



(16.222) 



dx 2 + dy 2 = dx /2 + dy'\ 



SEC. 16.2] THE SPECIAL THEORY OF RELATIVITY 489 

In fact, the quadratic expression dx 2 + dy 2 is an invariant. 
But in the space-time diagram 

dt 2 + dx 2 ^ dt' 2 + dx' 2 . 
The invariant quantity is, by (16.205), 

(16.223) dt* - dx 2 /c 2 = dt'* - dx' 2 /c 2 . 

In geometry, we denote the invariant (16.222) by ds 2 and 
call ds the distance between two adjacent points in the plane. 
This suggests that we should give a name to the square root of 
(16.223). However, the minus sign introduces a complication, 
since the invariant may be negative, and hence its square root 
imaginary. So we put 

(16.224) c ds 2 = dt 2 - dx 2 /c 2 , 

where e = +1 or 1, according as the expression on the right is 
positive or negative. We call ds the separation between the 
events (x } t) and (x + dx, t + dt). 

In the case of two events which are not adjacent, we define 
the separation as s fds, taken along the straight line joining 
the points in the space-time diagram which correspond to them. 
Since dx/dt is constant along a straight line, we easily find 

(16.225) es 2 = (t, - * 2 ) 2 - (x l - o- 2 ) 2 /c 2 , 

where the two events in question arc (xi, ti) and (z 2 , 2). We note 
that, if the two events have the same x, the separation is simply 
the difference between the ^s; if they have the same t, the separa- 
tion is the difference between the Z'H, divided by c. 

The lines in the space-time diagram satisfying one or other 
of the equations 

(16.226) dt = dx/c, dt = -dx/c, 

are called null lines, because the separation between any two 
points on such a line is zero. Clearly a null line represents 
the history of a flash of light traveling along the axis Ox of a 
Galilean frame, in one direction or the other. 

So far we have made no hypothesis regarding the motion of a 
particle in a Galilean frame. We shall postpone the discussion of 
motion under a force to Sec. 16.3, but now accept the law that a 
free particle travels in a straight line with constant speed, just as in 



490 



MECHANICS IN SPACE 



SEC. 16.2 



Newtonian mechanics. This means that a free particle traveling 
along Ox moves in accordance with 

dx 



where u is a constant. Its history appears in the space-time 
diagram as a straight line, with equation 



(16.227) 



t = constant. 

u 



We shall now show how the contraction of a moving rod and 
t 



t' 




Fio. 162. Space- time diagram for the 
contraction of a moving rod and the 
slowing down of a moving clock. 



* the slowing down of a moving 
clock appear in the diagram. 

In Fig. 162 the lines a, b 
represent the histories of the 
ends of a measuring rod lying 
on Ox' and fixed in S'. These 
lines are drawn parallel to !2', 
because the x 1 of each end of 
the rod is a constant. The 
length // (judged by S r ) is 
proportional to the separation 
A'B'. To get tin instantane- 
ous picture from the viewpoint 
of 8, we draw a line parallel 
to Six (i.e., with t constant), 
cutting a, 6 at A, B y respectively. Then the length L (as judged 
by S) is proportional to the separation AB. That L 5* L' is 
evident from the diagram. 

The line a in Fig. 162 may also be regarded as the history of a 
clock fixed in S'. In its history, A', A are two events, and the 
time interval T' between them (as judged by S') is the separa- 
tion A' A. As judged by 8, however, the interval between these 
events is T, the increase in t in passing from A 1 to A ; it is, in fact, 
the separation A'C, whore A'C is drawn parallel to Sit. It is 
evident that T f ^ T. 

There are some facts about the space-time diagram which we 
leave to the consideration of the reader. Why do the axes 
Qx'f not interlace with Slxtl Why does T appear smaller than 
T' in Fig. 162, whereas we have proved the reverse in (6.221)? 



SEC. 16.3] THE SPECIAL THEORY OF RELATIVITY 491 

Is it true, for a triangle in the space-time diagram, that the sum 
of the separations represented by two sides is greater than the 
separation represented by the third side? 

16.3. KINEMATICS AND DYNAMICS OF A PARTICLE 
Composition of velocities. 

Let P, Q be two particles traveling with uniform velocities 
MI, u z along the axis Ox of a Galilean frame of reference S. 
What is the relative velocity of the two particles? In Newtonian 
mechanics, we should answer: u z MI. In relativity, we say 
that this is only the difference between the velocities. The 
velocity of Q relative to P is the velocity of Q as estimated by a 
Galilean observer S' traveling along with P, i.e., using a Galilean 
frame of reference in which P is fixed. 

Between S and S r we have the Lorentz transformation 
[cf. (16.213)] 



(16.301) x' = yi (x - iii), f = 7i * - 

= _ _ 
71 vT^fA" 2 " 

Consider now the motion of Q; for it, dx/dt = u z , and its velocity 
as estimated by S f is 

/iAQH9\ / dx ' dx "" Ul di - u *~~ Ul 

(16.302) u = = 



dt _ Uldx/c * - i 
This is the law which replaces the Newtonian law, 
(16.303) u' = Ma - UL 

We note that, if u\ and i/ 2 are small compared with c, (16.302) 
differs very little from (16.303). 

If we solve (16.302) for u z and then make a change in notation, 
we get the rclativistic law of composition of velocities: If a particle 
moves with velocity HI in S', and S' has a velocity M 2 relative to S, 
then the velocity of the particle relative to S is 



Proper time. 

Let S be a Galilean frame of reference, and let P be a particle 
traveling along the axis Ox with uniform velocity M. It may be 



492 



MECHANICS IN SPACE 



[SEC. 16.3 



regarded as a particle of a second Galilean frame S r . Consider 
two adjacent events in the history of P; the time interval dt' 
between them (as estimated by a clock carried with P) is, by 
(16.213), 



dt' 



dt udx/c 2 

' 



But dx/dt = w, and so 

dt' = dt \A ~ u 2 



\/dt 2 - dx*/c* = ds, 



where da is the separation between the two events. Thus the 

separation equals the time interval, as measured by a clock carried 

with the particle. 

Hitherto we have considered only particles with uniform veloc- 
ities. We now think of a particle, 
traveling with accelerated motion. 
Its history appears in the space-time 
diagram as a curve. How does a 
clock behave if carried with the 
accelerated particle? Our previous 
hypotheses do not tell us; we must 
make a new assumption. This 
assumption is as follows: The time 
interval between two adjacent events in 
the history of an accelerated clock is 
given by the separation between these 
events. 
This separation is called the interval of proper time between 

the events. Thus the element of proper time for a particle 

moving along Ox with speed u is 




n / x 

FIG. 163. Space-time dia- 
gram of the histories of two 
clocks. 



(16.305) 

where 

(16.306) 



ds = dt/y u , 

1 



VT^^ 



Now we can give the explanation of the curious prediction 
made early in the chapter regarding the behavior of a clock taken 
on a flight. Figure 103 is a space-time diagram; a is the history 
of the clock that stayed at home. At the event A(t = ti) the 



SBC. 16.3] THE SPECIAL THEORY OF RELATIVITY 493 

other clock left and described the space- time curve 5, returning 
at the event B(t = t z ). Suppose both clocks read zero at A. 
Then at B the clock that stayed at home reads (since dx = 
throughout its history) 

(16.307) T = f* ds (along a) 

-JO 



The clock that flew reads 

(16.308) r = JT* ds (along 6) 



Thus 7" < T 7 , which proves the validity of the prediction. 

Equations of motion in absolute form. 

It has been remarked that gravitation lies outside the scope of 
the special theory of relativity. But there are available other 
forces by means of which accelerated motion may be produced. 
What we shall have to say is theoretically applicable to the 
accelerated motions of ordinary life, but the differences between 
the relativistic and the Newtonian predictions are then far too 
small to measure. The differences become appreciable only 
in the dynamics of atomic particles accelerated by forces of 
electromagnetic origin. However, the same principle applies 
throughout, and we may understand it by thinking of any 
small body under the influence of any force. 

We have accepted the hypothesis that all Galilean frames 
are equivalent. Thus, whatever form of equations of motion 
one Galilean observer adopts, a similar form must hold for any 
other Galilean observer. In fact, the equations of motion of a 
particle must be invariant under the Lorentz transformation. 

If we take the Newtonian equation 

(16.309) m S = P 

and apply the Lorentz transformation (16.213) to get an equation 



494 



MECHANICS IN SPACE 



[SBC. 16.3 



in x' and ', we find an equation of quite different form. Thus 
(16.309) is not invariant under the Lorentz transformation. 

To see what form of equation is suitable, we have to consider 
space-time vectors. 

In Fig. 164, we have taken a point A in the space-time diagram 

and drawn a directed segment SlA. This is a space- time vector; 
its components in the directions &r, 
tit are x, t, the space-time coordinates 
of A. If we use other axes toY, the 
same vector has different components. 
But between the two sets of com- 
ponents the Lorentz transformation 
holds. 

We define a space-time vector as 
a pair of quantities (J, r) which trans- 
form, when we change axes in space- 
time, just like (x, t), i.e., according to 




A space-time vec- 
tor. 



(16.310) 



Vi - v-/c 2 ' 

As we have stated, our problem is to build equations of motion 
invariant under a Lorentz transformation. The key to the 
solution is found in the idea of the space-time vector. We shall 
form equations in which a space-time vector is equated to a space- 
time vector. 

Consider a particle moving along Ox with a general motion. 
This corresponds to some curve in space time. The proper time, 
measured from some initial point, may be taken as a parameter, 
and the equations of the curve written 

x = X ( S ), t = t(s). 

Since ds is an invariant, the pair of quantities (dx/ds, dt/ds) 
is a space-time vector. We see this by differentiating (16.213). 
We call (dx/ds, dt/ds) the absolute velocity of a particle. Explic- 
itly we have, by (16.305), 



(16.311) 



dx 
ds 



dt 



SBC. 16.3] THE SPECIAL THEORY OF RELATIVITY 495 

where u = dx/dt, the velocity of the particle relative to the 
Galilean frame of reference S, corresponding to Slxt. If the 
particle has a velocity small compared with that of light, so that 
u/c is small, the components of the absolute velocity are approxi- 
mately (u, 1). 

Similarly, (d 2 x/ds 2 , dH/ds 2 ) is a space-time vector. We call 
it the absolute acceleration. 

We now accept, as satisfactory from the point of view of 
invariance under the Lorentz transformation, the following 
equations of motion: 

t^T rl^t 

(16.312) m ^ = X, m,4i=r, 

as as 

where mo is a constant (the proper mass of the particle) and 
(X, T) is a space-time vector, called the absolute force. 

In a different Galilean frame of reference $', with velocity V 
relative to $, these equations read 



where 

(16.313) X' = y(X - VT), T' = y (T - Y 

= 1 

7 VT^TVc 2 " 

This may be verified immediately by applying (16.213) to 
(16.312). 

Equations of motion in relative form. 

Let us now put the equations of motion (16.312) into another 
form in order to show the relation of relativistic to Newtonian 
mechanics. Since ds = dt/y u , these equations may be written 

(16.314) j t (m y u u) - X/y uy ~ (m T) = T/y u . 
Let us define some terms, as follows: 



(16.315) 



Relative mass = m m Q y u = 



VI ~ u 2 
Relative momentum = mu. 



Relative force = P = X/y* = X 
Relative energy = E = me 2 



496 MECHANICS IN SPACE [Sue. 16.3 

Then the first of (16.314) may be written 

(16.316) ~ (mi*) = P; 

in words, 

rate of change of relative momentum = relative force. 

This is the equation of motion of a particle moving on the z-axis 
under the influence of a force P. It is of the Newtonian form, but 
with a remarkable difference. The (relative) mass of a particle 
is not a constant; it varies with the speed of the particle according 
to (16.315). 

If u/c is small, the variation of m from the value m is insignifi- 
cant, but on the other hand m tends to infinity as the speed of the 
particle (u) tends to that of light (c). No particle has ever been 
observed traveling with a speed equal to, or greater than, that of 
light. This physical fact agrees with the theory. If the speed 
of a particle were to increase up to and beyond the speed of light, 
the relative mass would become meaningless, passing through an 
infinite value to imaginary values. 

To discuss the second equation of (16.314), let us first return to 
(16.224). This may be written 

(16.317) 

since (dx/dt) 2 < c 2 for a particle. Differentiation gives 



na<Mtt dt dH 

( lO.oIo) i ; 5 5 -7 r~5 =s u - 

N ds ds 2 c 2 ds ds 2 

Thus, by (16.312), 

(16.319) T ~ - \ X ~ = 0, 

ds c 2 ds 

or 

(16.320) c 2 T = Xu = Puy u . 

In fact, the second component of absolute force is closely related 
to the first component. 

If we multiply the second of (16.314) by c 2 and substitute from 
(16.315) and (16,320), we get 

(16.321) -jj- = Pu. 



SEC. 16.3] THE SPECIAL THEORY OF RELATIVITY 497 

This is the equation of energy and justifies the definition of relative 
energy as in (16.315). For (16.321) reads, in words, 

rate of change of relative energy 

= rate of working of relative force. 

This has the Newtonian form, but the expression for energy (E) 
does not at first appear related to the Newtonian kinetic energy. 
However, if we expand by the binomial theorem, we obtain 

< 16 - 322 ' B 



and if u/c is small, we have approximately 

(16.323) E = m r 2 + %m<>u*. 

This differs from the Newtonian expression for kinetic energy 
only by the constant m c 2 , which is called the rest energy or 
proper energy. 

The quantity Woe 2 appears of little importance here, because 
E is differentiated in (16.321), and so the constant disappears. 
The equation (16.321) would still hold if we had adopted the 
definition 

(16.324) E = . m c2 ..- - m c 2 



for relative energy. There are, however, good reasons for 
preferring (16.322) to (16.324) as a definition of energy. Some of 
these are connected with the disintegration of atoms and their 
structure. The known atomic weights of the elements* are con- 
sistent with the principle of energy only if (16.322) is regarded 
as the energy of a particle. In relativity, mass and energy are 
no longer distinct concepts. Even when a particle is at rest, it 
has energy m c 2 , and wo cannot convert this energy into another 
form without destroying or altering the mass m . 

Example. Consider a particle moving on the x-axis under a constant 
relative force P, starting from rest at the origin at t = 0. By (16.316) the 
motion satisfies 



498 MECHANICS IN SPACE [SEC. 16.4 

Hence, 

(16.326) , WoU - Pt. 



Solving for w, we get 

(16.327) u cPt 



4- wjc* 

We note that u is less than c for all values of t and tends to the limiting value 
c as t tends to infinity. This behavior is, of course, quite different from the 
behavior of a particle under constant force in Newtonian mechanics. 
Since u = dx/dt, (16.327) gives, on integration, 



(16.328) x 

If m c is large compared with Pt, this reduces approximately to 

(16.329) x - * * 2 , 
the familiar Newtonian formula. 

16.4. SUMMARY OF THE SPECIAL THEORY OF RELATIVITY 

I. There is no such thing as absolute time. 

II. Lorentz transformation : 

(16.401) x 1 = y(x - Vt), t' = 7 ft - ~ 
7 



III. Space -time diagram. 

(a) An event is represented by a point. 

(6) The history of a free particle is represented by a straight 
line. 

(c) The history of a flash of light is represented by a null line. 

(d) The separation of two adjacent events is ds, where 

(16.402) 6 ds 2 = dt z - dx*/c* (t = 1). 

IV. Kinematics and dynamics of a particle. 

(a) Element of proper time for moving particle: 



(16.403) ds = dt 



Ex. XVI] THE SPECIAL THEORY OF RELATIVITY 499 

(6) Equation of motion: 

(16.404) (mow) = P, 7* 



(c) Energy: 

(16.405) E = rao7t*c 2 = m c 2 + w M 2 , approximately. 

(16.406) ^ = P^ 



EXERCISES XVI 

1. An airplane sets out to fly at 500 miles per hour. Show that it would 
have to fly for more than a thousand years in order to make a difference 
of one one-hundredth of a second between the times recorded by a clock in 
the airplane and a clock on the ground. 

2. Show that, if x/c and t are taken as coordinates in the space-time 
diagram, the history of a flash of light is equally inclined to the axes. Draw 
the history of a flash which passes to and fro between a mirror fixed at the 
origin and a mirror which moves along the observer's axis Ox with constant 
speed. 

3. Prove directly from the formula (16.304) that, if the magnitudes of 
u\ and u z are both less than c, the magnitude of the velocity relative to S is 
less than c. 

4. Two electrons move toward one another, the speed of each being 0.9c 
in a Galilean frame of reference. What is their speed relative to one 
another? 

5. For suitably chosen axes in two Galilean frames S and S r , the com- 
plete Lorentz transformation is 

x' = y(x - Vt), y f =y, z' = z, t' = y(t - Vx/c*), 



where V is the relative velocity of S and S f . 

A particle, as observed by 5', describes a circle x' z + y'* = a 2 , z' = 0, with 
constant speed. Show that to S the particle appears to move in a ellipse 
whose center moves with velocity V. 

6. All electromagnetic waves travel with the fundamental velocity c in 
empty space. A radio station fixed in a Galilean frame of reference S sends 
out waves. Show that, to an observer in another Galilean frame S', these 
waves at any instant form a family of nonconcentric spheres. Is it possible 
that two of these spheres should intersect? 

7. Show that the Lorentz transformation may be regarded as a rotation 
of axes through an imaginary angle. 

8. The history of a moving particle is represented in the space-time 
diagram by the hyperbola 



500 MECHANICS IN SPACE [Ex. XVI 

Show that 

d*x 19 dH 19 . 

-3-^ = & 2 , -r-i - Wt. 

ds 2 ' ds* 



9. Two particles, with proper masses mi, m 2 , move along the axis Ox 
of a Galilean frame with velocities Ui, M 2 , respectively. They collide and 
coalesce to form a single particle. Assuming the laws of conservation of 
relativistic momentum and energy, prove that the proper mass m* and 
velocity MS of the resulting single particle are given by 



ml 



f 1 -- ^ 

iUi -f ?n 2 72i/2 



Wl7l T" W272 

where yf = 1 - w?/c 2 , 7^ = 1 - u\/c*. 

10. A particle of proper mass m moves on the axis Ox of a Galilean frame 
of reference, and is attracted to the origin by a (relative) force m^x. It 
performs oscillations of amplitude a. Show that the periodic time of this 
relativistic harmonic oscillator is 



where 



Verify that, if c > , r * 2ir/fc (the Newtonian result) ; and show that if 
ka/c is small, 



27r A fc2o 2 \ . 

j ^1 + A -~T )' approximately. 



APPENDIX 
THE THEORY OF DIMENSIONS 

Two physicists are shipwrecked on a desert island. After 
making qualitative observations of their surroundings, they wish 
to make measurements. But here a difficulty arises, for they 
have none of the usual apparatus of the laboratory no meter 
scale, no set of weights, no clock.* Everything they require 
they must construct for themselves. 

If they can agree on the length of a certain stick as unit of 
length, the mass of a certain stone as unit of mass, and the dura- 
tion of some simple rcpeatable experiment as unit of time, all will 
be well; both experimenters will assign the same number to the 
same measurable quantity. But why choose one stick rather 
than another, one stone rather than another, one experiment 
rather than another? If the two physicists are obstinate, each 
in his own preference of units, there is no valid argument by 
which one can persuade the other to yield. 

This disagreement concerns only physical measurements. In 
the realm of pure mathematics, there is complete accord; both 
agree, for example, that 

(1) 2 + 2-4, (x + l)(s - 1) - x 2 - 1, 



But in the matter of the choice of units, we may well imagine 
that neither physicist will yield to the other. So they decide 
to work independently, each constructing his own apparatus, 
measuring quantities in the units he prefers, and developing his 
own results. If they wish to discuss their work, how is one to 
interpret the results of the other? How far do their individual 
efforts contribute to the construction of a common science, 
independent of the choice of units? These are questions which 
belong to the theory of dimensions. 

* We may suppose the sky perpetually overcast, so that the rotation of the 
heavens cannot be used as a clock. 

501 



502 PRINCIPLES OF MECHANICS 

It may appear strange that we have been able to postpone to 
an appendix the discussion of these important questions. The 
explanation is that the theory of dimensions becomes necessary 
only when we wish to compare results for two different systems 
of units. In our work, we have used arbitrary units and letters 
(algebra) instead of actual numbers (arithmetic). Our results 
are valid quite generally and can immediately be applied in any 
particular system of units. 

Perhaps an analogy with analytical geometry will be helpful. 
We may develop results true for arbitrary Cartesian axes, e.g., 
properties of conies deduced from a general equation of the 
second degree. We meet the theory of transformations (the 
analogue of the theory of dimensions) only when we consider two 
different sets of axes at the same time. The invariants of analyti- 
cal geometry are analogous to general physical laws, true for all 
systems of units. 

Units and dimensions. 

The theory of dimensions arises from the fact that units may 
be chosen arbitrarily. Instead of arguing over the respective 
merits of different systems of units (e.g., centimeter-gram-second 
and foot-pound-second), let us regard all systems as equally 
valid. Each physicist may select his own units. This means 
that he selects a piece of matter and says that its mass is unity, 
he selects a rigid bar and says that its length is unity, and he 
selects a repeatable experiment and says that its duration is 
unity. He can now measure masses, lengths, and times, and 
record them as so many units. To find a velocity, he measures 
distance traveled and time taken, and divides the one number 
by the other. He deals similarly with acceleration, moment of 
inertia, kinetic energy, and so on. 

As for force, there are two possible plans: (i) he may use 
Newton's law of motion in the form P = mi to define force in 
terms of mass and acceleration, or (ii) he may use a separate 
arbitrary unit of force. The second plan is good in statics, but 
the first is far simpler in dynamics and may be used in statics 
also. We shall accept the first plan for the present discussion. 
With this understanding, all the quantities occurring in mechanics 
are built up out of mass, length, and time; the physicist can assign 
numerical values to them all, once he has selected his funda- 
mental units of mass, length, and time. 



THEORY OF DIMENSIONS 503 

Two physical quantities may have different numerical values 
and yet be of the same type. For example, the linear momenta 
of two particles may have different numerical values, but they 
are both built up out of mass, length, and time in the same 
definite way; in fact, 

(2) linear momentum = masa * length . 

time 

To express this more compactly, we introduce the symbols 
M , L, T for mass, length, and time, and write symbolically 

(3) [linear momentum] = [MLT' 1 ]. 

The square brackets are to remind us that this is no ordinary 
equation connecting numbers but a symbolic shorthand to show 
how linear momentum involves the fundamental quantities. 
This is called the dimensional notation; we say that linear 
momentum "has the dimensions [MLT~ 1 }" All the quantities 
occurring in mechanics may be expressed dimensionally in the 
form 



where a, /3, 7 are positive or negative powers, not necessarily 
integers. The following list of dimensions is easily verified: 

[velocity] = [LT~ l ], 

[acceleration] = [LT~*\, 

[force] = 

[moment of a force] = 

[linear momentum] = [MLT~ l ], 

[angular momentum] = [ML 2 jT -1 ], 

[energy] = [ML*T~*], 

[angular velocity] = [T 7 " 1 ]* 

[moment of inertia] = [ML*]. 

In writing down the dimensions of a physical quantity, we pay 
no attention to numerical factors. Thus the dimensions of 
and mf are the same, viz., [ML 2 T~ 2 ]. 

We do not add or subtract quantities having different dimen- 
sions, but we frequently multiply such quantities by one another 
or divide them by one another. The rule by which we obtain the 



504 PRINCIPLES OF MECHANICS 

dimensions of the product or quotient is obvious from the defini- 
tion of dimensions. It is as follows: 
Let Qi and Q 2 be physical quantities with dimensions 



[Qi] 
then 



It AHSxaj 

te] 

Led 



If Qi and Q 2 have the same dimensions, then 



and we say then that Qi/Qz is dimensionless. For example, the 
circular measure of an angle is obtained by dividing a length 
(arc) by a length (radius), and so an angle is dimensionless. 
It is easy to verify that the following combinations are also 
dimensionless: 

force X time 






linear momentum 
force X length 

energy 

moment of inertia X angular velocity 
angular momentum 

Exercise. Einstein's radiation formula is E =* hv, where E is the energy 
of a photon, v is its frequency, and h is Planck's constant. Show that 
Planck's constant has the dimensions of angular momentum. 

Change of units. First method. 

Let us now consider two physicists Si and $ 2 , who use different 
units of mass, length, and time. When they measure the same 
physical quantity, they record different results. But, as we shall 
now see, it is easy to pass from one numerical value to the other 
when we know the ratios of the two sets of units. 

For symmetry, we introduce a third physicist $o, using a 
third system of units; we shall call his units "absolute" for 
purposes of reference, without meaning to imply that they are 
in any way more fundamental than the units of Si or # 2 . Let 
the units of *Sfi contain m\ t l\, and h, absolute units of mass, length, 



THEORY OF DIMENSIONS 505 

and time, respectively; and let the units of $2 contain w 2 , 1%, and 
tz absolute units. 

Consider a physical quantity Q with dimensions [M*LPTi\. 
This quantity is measured by S Q , Si, and S z , with numerical 
results as follows: 

/So Oi >32 

Qu Qi Q*. 

Now every unit of mass recorded by Si corresponds to mi absolute 
units, and similarly for length and time. Hence, one /Si-unit 
of the quantity measured corresponds to m-flfti* absolute units, 
and Qi 5i-units correspond to Qimi a li^ absolute units. But we 
know that Qi /Si-units correspond to Qo absolute units, and so 

(4) Qo = 
similarly, 

(5) Qo = 
i 

Comparing (4) and (5), we see that the law of transformation 

connecting the results of Si and $ 2 is 

(6) 
or 

m 
(7) 

If we identify the units of S with those of Si, so that the 
absolute units are now the /Si-units, we have 

mi = 1, Zi = 1, ti = 1, 
and so 



m 

where mz, h, fa are the numbers of $i-units contained in the 
$ 2 -units. This formula gives the number Q 2 assigned by <S> 2 to a 
quantity, in terms of the number Qi assigned by Si to the same 
quantity and the ratios of the units. 

Exercise. An energy is 362 in c.g.s units. What is its numerical value 
in f.p.s. units? 



506 PRINCIPLES OF MECHANICS 

Change of units. Second method. 

The above method is logical, but not good in practice. Con- 
version from one set of units to another is a process which we 
must be able to carry out quickly and accurately, and the rules 
should be simple and easy to remember. The formula (7) is bad 
because it involves a third system of units, and (8) is bad because 
it is unsymmetrical and hard to remember. The method we are 
about to describe is that in common use. 

Compare the equations (1) with the following: 

(9) 1 meter = 100 cm., 1 Ib. = 453.6 gm., 

22 ft. per sec. = 15 miles per hr. 

These are true statements, but they differ from (1) in an impor- 
tant respect: the equations (1) involve only pure numbers, 
whereas (9) involve measurable physical quantities. To dis- 
tinguish them, we may call (1) mathematicians' equations (or 
briefly M-equations) and (9) physicists' equations (or P-equa- 
tions). We know what we can do with M-equations according 
to the methods of algebra and calculus. There are certain rules 
of manipulation, which we apply with confidence that we shall 
never reach a false conclusion. Let us boldly apply the rules of 
algebraic manipulation to P-equations, treating such words as 
meter, cm., Ib. as if they were ordinary algebraic symbols. A 
word of warning, however the signs =, +, and are to be 
used to connect only quantities of the same type, i.e., of the same 
dimensions. 

We think again of two physicists Si and $2. Let Si name his 
units Mi, Li, Ti] and let S z name his units M 2 , 1/2, T z . These 
are names (like gm. or cm.), not numbers. If Si measures a 
length, he records the result in the form 

Q = QiLi; 

this is a P-equation, in which Q stands for "the quantity which 
is measured," arid Qi is a number. More generally, if Si meas- 
ures a quantity with dimensions [M a UT^] t he records 

(10) Q = QiMfLJTf, 

where Q\ is a number. If Sz measures the same quantity, he 
records 

(11) Q 



THEORY OF DIMENSIONS 507 

There is nothing novel about this; it is what we do when we write 

acceleration due to gravity = 32 ft. sec.~ 2 
acceleration due to gravity = 980 cm. sec.~ 2 

Now we bring into operation our assumption that the P-equa- 
tions (10) and (11) may be treatod in the same way as we should 
treat M-equations. We get at once 

(12) QJlfLJTi* 

and 



If we interpret Mi/Mz to mean the ratio of the unit M i to the 
unit Mz, then Mi/Mz is a pure number in fact, the measure 
of Mi in terms of M z . Since Li/Z/2 and T\/T* may also be 
regarded as pure numbers, (13) is an M-equation, although (12) 
(from which it was obtained) is a P-equation. 

Equation (13) is what wo have been seeking a formula to 
give Qz when Qi and the ratios of the units are known. If we 
lack confidence in it, because it has been obtained by a symbolic 
method, we can reassure ourselves by turning back to the first 
method; there only M-equations were used, and the deduction 
of (6) and (7) is logically sound. We see that (12) is merely the 
P-equation corresponding to the M-equation ((>), and (13) is the 
same as (7) both M-equations. 

When we actually carry out a conversion from one system 
of units to another, it is the P-equation (12) rather than the 
M-equation (13) that we use. It would, however, be more 
correct to say that we use neither. Therein lies the simplicity 
of the symbolic method; we treat each problem on its merits, 
without having to remember anything, except that it is permis- 
sible to use the symbolic method, in which words are treated as 
algebraic symbols. The formulas (12) and (13) were obtained 
only for purposes of comparison with (6) and (7). 

To show the symbolic method in action, let us convert an 
acceleration of 32 ft. sec." 2 into mile hr.~ 2 * First we write down 

* It is convenient to write each unit in the singular, to avoid waste of 
energy in deciding whether to use the singular or the plural. This is a 
mathematical symbolism, and in it simplicity is more important than 
grammar. 



508 PRINCIPLES OF MECHANICS 

1 mile = 5280 ft., 1 hr. = 3600 sec., 



so that 



1-1 1 1 ^ 

lsec.=^hr. 



Then, 

(1 ft.) 



32 ft. sec.~ 2 = 32 



(1 sec.) 2 



= 32^ 



^3600 
32 X 3600 X 3600 ., , _ 2 

5280 milehr * 

= 78,545fV mile hr.~ 2 

A physicist would round off the result. For he would think of 
the number 32 as obtained by measurement carried out only to 
two-figure accuracy, and so he would prefer to write 

32 ft. sec." 2 = 79,000 mile hr.~ 2 

This question of "significant figures" has nothing to do with the 
theory of dimensions, and we shall not pursue it further. The 
discrepancy between the two statements arises from the two ways 
of thinking mathematical and physical which we mentioned 
in Chap. I. 

Exercise. Work out the exercise on page 505 by the above method. 

Dimensionless quantities and physical laws. 

If a quantity is dimensionless, then a = j3 y in (13), 
and therefore Q\ = Q 2 . A dimensionless quantity has a value 
independent of the system of units employed. This fact makes 
such quantities particularly simple to handle, because any 
possible confusion regarding units is automatically eliminated. 
Any mathematical combination of dimensionless quantities is 
itself dimensionless. 

We can now answer the questions raised in connection with the 
two shipwrecked physicists. The formulas given above enable 
the one to interpret the results of the other, i.e., to transform them 
into his own units. As for the second question the building up 
of a common science independent of the choice of units the 



THEORY OF DIMENSIONS 509 

answer is to be found in the concept of the dimensionless quan- 
tity. Any equation connecting dimensionless quantities is true in 
all systems of units, if true in one. 

Suppose, for example, that a physicist (prior to the time of 
Galileo) made measurements on a falling body, using some 
system of units of length and time. We assume that he was able 
to measure the height h from which the body fell, the speed q 
with which it struck the ground, and the time t it took to fall. 
Suppose he found 

2? -2 

h ~ 2j 

for a whole set of experiments in which h was given different 
values. He would have been justified in regarding this as a 
result of great importance, because it holds in all systems of units, 
since qt/h and the pure number 2 arc both dimensionless. 

As shown above, any equation connecting dimensionless 
quantities is a physical law, in the sense that its truth is inde- 
pendent of the choice of units. However, it is not necessary 
to express a physical law in dimensionless form. It is merely 
necessary that the equation should be dimensionally homo- 
geneous; i.e., the terms equated to one another must have the 
same dimensions. This will ensure that the law is true in all 
systems of units, if true in one. 

The constants occurring in physical laws usually have dimen- 
sions. Consider the law of gravitational attraction (6.501) 

n Gmm' 



where P is the magnitude of the force between particles of 
mass m, m' at a distance r apart. To make this dimensionally 
homogeneous, we must assign suitable dimensions to the con- 
stant G. This is easily done if we write the equation in the form 



mm 
We have then 



[01- 



= [M- 1 L*T- 2 ]. 



510 PRINCIPLES OF MECHANICS 

In the c.g.s. system, 

G = G.67 X 10~ 8 gm.- 1 cm. 3 sec.~ 2 

Exercise. Form a dimensionless combination of the gravitational con- 
stant, density, and time. 

Applications. 

Apart from its use in the change of units, the theory of dimen- 
sions has three important applications: 

(i) It supplies us with a useful check against slips in calculation. 

(ii) It suggests forms of physical laws. 

(iii) It enables us to predict the behavior of a full-scale system 
from the behavior of a model. 

These applications will now be explained. 

(i) Provided that we do not insert numerical values, the dimen- 
sions of every combination of symbols occurring in our work 
are obvious. For example, if a is the length of a pendulum and 
g the acceleration due to gravity, then 



The basic law of motion (1.402) is dimensionally homogeneous, 
in the sense that both sides have the same dimensions, viz., 
[MLT~ 2 ]. The operations we perform on this equation may 
change the dimensions of the two sides, but they are both changed 
in the same way. Thus, at all stages of our deductions we have 
dimensionally homogeneous equations. Indeed, it is inevitable 
that this should be so, since otherwise the two sides of an equa- 
tion would change differently on change of units, and if true for 
one system of units would not be true for another. This gives 
a useful check. For example, suppose we are engaged in working 
out the formula for the periodic time of small oscillations of a 
simple pendulum. As a result of our work we arrive, perhaps, 
at the result 

r = 2*. 

g 

Dimensionally, this reads 

[T] = m 

which shows that the result is incorrect. Such a check will, of 
course, never be of any assistance as far as a numerical coefficient 



THEORY OF DIMENSIONS 511 

is concerned; for example, the theory of dimensions alone cannot 
tell us that 



is incorrect. 

Exercise. It is suggested that the equation of motion of a particle on a 
line is 

d*x , dx 



where a has the dimensions [L], and 6 the dimensions [jT~ 2 ]. Would you 
accept this equation as correct? 

(ii) To see how the theory of dimensions suggests forms of 
physical laws, we shall consider the transverse vibrations of a 
heavy particle at the middle point of a stretched string. The 
quantities involved are 

ra = mass of particle, 
a length of string, 
S = tension, 
T = periodic time. 

The periodic time must be some function of the quantities m, a, 
S, and so we write 

T = /(m, a, S). 

The only combination of m, a, S having the dimensions [T] 
is of the form Cm a a (3 S y , where C, a, 0, y are pure numbers, at 
present unknown. Accordingly we assume 



and obtain the dimensional equation 
[T] = [ 



Hence, 

tt + 7 = 0, /3 + 7 = 0, -27 = 1, 
or 

and so our formula for T is 

Ima 



512 PRINCIPLES OF MECHANICS 

We cannot find the numerical factor C from the theory of dimen- 
sions. To obtain it theoretically, we must solve the differential 
equation of motion. But, if we are satisfied with an experimental 
result, one experiment will suffice to determine C. 

This method is useful in the case of a complicated system, 
where the direct solution of the differential equations is difficult. 

Exercise. Consider the transverse vibrations of a system consisting of 
20 equal particles, equally spaced on a stretched string, Show that each 
of the twenty normal periods is of the form 



r> \ 

r = C V-^, 

where m is the mass of each particle, a the length of the string, S the tension 
in it, and C a numerical constant which may depend on the particular 
normal mode. 

(iii) To show how the theory of dimensions enables us to use a 
model to predict full-scale phenomena, let us consider the flow 
of air past the wing of an airplane. The lift Y on the wing 
obviously depends on the following quantities: 

p the density of the air, 

U = the speed of the wing relative to the air, 
I = a linear measurement of the wing (e.g., its width from 
back to front, at some definite position). 

The lift depends, of course, on the shape of the wing; we shall 
consider only wings of one definite shape, transformed into one 
another by changing the length I. 

The problem is to calculate the lift Y on the full-scale wing 
from the measurement of the lift Y' on a model. Now Y is a 
function of p, U, 1] and Y' is the same function of p', U f , /', where 
the accented quantities refer to the experiment on the model, 
the same units of mass, length, and time being used in both 
cases. So we write 

F = f( P , U, 1), Y' = f( P ', U', I'). 
As in the preceding example, we take 

/(p, U, I) = CfV'f, 

where C is a pure number. Since [p] = [ML" 8 ], [U] = [LT~ l ], 
[1] = [L], and [F] = [MLT~*], we easily find 

Y = CpUH\ Y' = Cp'U'H' 2 . 



THEORY OF DIMENSIONS 513 

Hence the full-scale lift is 



If the density of the air is the same for both cases, this becomes 



772/2 

V V 



.. 72 
y i/, 



When we insert the numerical values for 7', /', Z', obtained from 
experiment on a model in a wind tunnel, and the values of U and 
/ appropriate to the full-scale wing in flight, we are able to read 
off the value of the lift F. 

Exercise. In order to study the (Inflections in an elastic beam with con- 
tinuous and isolated loads (cf . Sec. 3.3), an engineer builds a model of the 
same material with a linear ratio 1 : 100. Show that, if the deflections in the 
model are to be one one-hundredth of the full-scale deflections, the con- 
tinuous load per unit length in the model-must be one one-hundredth of the 
full-scale loud. In what ratio should the isolated loads be reduced? 



INDEX 

The numbers in heavy type refer to the Summaries at the ends of the 
chapters. 



Absolute equations of motion, 493- 

495 
Absolute velocity, acceleration, and 

force, 494, 495 

Acceleration, 27, 28, 36, 305, 332 
absolute, 495 
of automobile, 205, 206 
complementary, 349 
of Coriolis, 349 

in cylindrical coordinates, 306, 307 
due to gravity (see </) 
radial and transverse components, 

120, 126 

in relativity, 492, 493, 495 
in spherical polar coordinates, 407, 

468 

tangential and normal compo- 
nents, 118, 119, 126, 305, 306, 
332 

of transport, 349 
Accelerations, composition of, 141, 

307, 308 

Action and reaction, law of, 32, 36 
Addition of vectors, 19-22, 36 
Air, resistance of, 151, 154-159, 184 
Airplane, 271, 272, 444, 512, 513 
Ames, J. S., 16 
Amplitude of oscillations, 162 
Angle of friction, 87, 88, 114 
Angular impulse, 357 
Angular momentum, in impulsive 
motion, 229-231, 238, 357, 358, 
361 

of particle, 128, 129, 146, 329, 333, 
341, 360 



Angular momentum, relative to 
mass center, 135, 136, 147, 330, 
345, 355, 360, 361 
Angular momentum, of rigid body, 
193, 194, 196, 222, 223, 330-332, 
333 
of system, 134-136, 147, 329, 330, 

344, 345, 360 

Angular velocity, of the earth, 143 
of rigid body, 122, 126, 308-311, 

332 

Anomaly, 187 
Aphelion, 180 
Appell, P., xi 
Applications, in dynamics in space, 

364-411-414, 418-463, 464 
of Lagrangc's equations, 467-472 
in plane dynamics, 151-184-186, 

189-222, 223 

in piano statics, 74-113-115 
in statics in space, 275-278, 296- 

301 

Applied force, 58, 295 
Approximations for electromagnetic 

lenses, 397-403 
Apse, 171-174, 186 

advance of, for spherical pendu- 
lum, 381 

Apsidal angle, 173, 174, 378-381 
Archimedes, 82 
Areal velocity, 170, 180, 186 
Associative property of vector addi- 
tion, 22 

Astatic center, 73 
Astronomical frame of reference, 31, 

33, 141, 143 
Astronomical latitude, 145, 403 



515 



516 



INDEX 



Attraction, electrostatic, 176 
gravitational, 82-86, 114, 144- 

146, 176, 177, 404, 405, 509 
Automobile, 204-206 
Axes of inertia, principal, 316-324, 

333 
Axially symmetric electromagnetic 

field, 387-403, 413 
Axis, of rotation, instantaneous, 308 
of screw displacement, 285 
of symmetry, 78, 321, 322 
of wrench, 269, 270 



B 



Balancing, problems of, 219-222 
Ball slipping on table, 447-450, 464 
Ballistic pendulum, 235, 236 
Ballistics, 151, 184 

(See also Projectile) 
Bars in frame, 106, 108 
Base point, 61, 259, 280, 281 

change of, 260, 283, 284 
Beam, internal reactions in, 92, 93, 

114, 272, 273 
thin, 92-98, 114 
Becker, K., 151 
Bending moment, 92-98, 114, 272, 

273 

Billiard ball, 447-450, 464 
Binormal, 264 
Blank, A. A., 388 
Body centrode, 124, 126, 309 
Body cone, 309, 421, 425, 427, 463 
Bound vector, 18, 19 
Bridge, suspension, 100, 114 



Cable, flexible, 98-105, 114, 116 
in contact with curve, 104, 105, 

116 
in space, 265, 266 

Cajori, F., 32 

Calibration of spring, 17 

Campbell, J. W., 103 

Cardan's suspension, 418 



Catenary, 100-104, 116 
Celestial pole, motion of, 428, 429 
Center, astatic, 73 
of gravity, 84-86, 114, 271 
instantaneous, 123-126 
of mass (see Mass center) 
of oscillation, 201 
of percussion, 2^9 
of system of parallel forces, 271 
Centimeter, 13 
Central force, general, 128, 168-176, 

186 
varying directly as distance, lb'8, 

169, 186 

varying as inverse square of dis- 
tance, 176-184, 186, 462 
Central symmetry, 77 
Centrifugal force, 143-145, 147, 349, 

350, 406 

Centrode, 124, 126, 309 
Chain (see Cable) 
Chako, N., 388 
Change of base point, 260, 283, 284 

of units, 504-508 
Charge on electron, 386, 387 
Charged particle, in axially sym- 
metric electromagnetic field, 
388-403, 413 
in electromagnetic field, 176, 381- 

403, 412, 413 
in uniform electromagnetic field, 

383-387, 412 
Chasles' theorem, 303 
Circular disk and cylinder, moments 
of inertia of, 190, 191, 222, 324 
Circular motion, 28 
Circular orbit, stability of, 174-176 
Clock, 12, 13, 501 
in relativity, 475-480, 482, 483, 

487, 490, 492, 493 

Clocks, synchronization of, 477-480 
en x, 368, 412 
Coefficient, of friction, 87, 88, 114 

of restitution, 232-234, 239 
Collar, A. R., 316 
Collisions, 231-235, 239 



INDEX 



517 



Commutative property in vector 

operations, 19, 21, 246, 257 
Complementary acceleration, 349 
Complex frame, 111, 112 
Components of vector, 22-24, 32, 

35, 36, 245 

(See also Acceleration; Velocity) 
Composition, of accelerations, 141, 

307, 308 
of couples, 2G8 
of finite rotations, 20, 282 
of infinitesimal displacements, 

281-283, 302 

of velocities, 140, 307, 308, 491 
Compound pendulum, 198-201, 223, 

370 
Compression, 233, 234 

modulus, 186 
Cone, body or polhode, 309, 421, 

425, 427, 463 
of friction, 87 
space or horpolhode, 309, 421, 425, 

427, 453 

Configuration of a system, 64, 286 
Conical pendulum, 374, 375 
Conjugate lines, 304 
Conservation of energy, 130, 131, 
137, 146, 147, 196, 201, 223, 231, 
341, 346, 356, 360, 361 
Conservative field, 66 
Conservative system, 65-67, 294, 

295, 302 

Constant of gravitation (sec Gravi- 
tational constant) 
Constraints, 57-60, 285-292 
moving, 474 
workless, 54-57, 70 
Contact, rolling, 55-57, 70, 123, 124 
rough, 86-88, 114 
smooth, 54, 55, 57, 70, 86 
Continuity of bodies, 14, 15, 76 
Contraction of moving rod, 486, 487, 

490 

Coordinate vectors, 23 
Coordinates, cylindrical, 306, 307, 

338 
generalized, 285-292, 302,462-472 



Coordinates, spherical polar, 467 
Coriohs, acceleration of, 349 

force, 143, 144, 147, 349, 406 
Coulomb's law, 176 
Couple, 49 

gyroscopic, 442-444, 464 

impulsive, 229 

moment of, 49, 50, 267 

twisting, in a beam, 272 

work done by, 64, 293 
Couples, composition of, 268 
Courant, R., 320 
Covering operation, 320 
Cranx, C., 151 

Critical form for a frame, 113 
Cuboid, moment of inertia of, 324 
Curvature, radius of, 119, 264 
Curves in space, 264, 265 
Cuspidal motion of a top, 436-438 
Cylinder, balancing problem for, 
219-222 

moments of inertia of, 191, 222, 
324 

rolling down inclined plane, 202- 

204 
Cylindrical coordinates, 306, 307, 338 

D 

Dale, J. B., 370 

D'Alembert's principle, 138, 139, 147 
Damped oscillations, 163-168, 185 
Deadbeat oscillations, 166, 185 
Decomposition, method of, 78, 79, 

114, 325 

Decrement, logarithmic, 166 
Degrees of freedom, 207, 208, 287 
Dcimel, R. F., 445 
Density, 76, 77 
Determination of past and future, 

340, 341 
Deviations due to earth's rotation, 

145, 407, 408, 414 
Differentiation, used to find moments 

of inertia, 325, 326 
of vectors and their products, 24- 
26, 35, 250, 257 



518 



INDEX 



Digonal symmetry, 321 
Dimensional notation, 503 
Dimensionless quantity, 504, 508 
Dimensions, theory of, 501-513 
Directed line, 22 
Discontinuity in bodies, 14, 15 
Discontinuous motion, 231 
Disk, moment of inertia of, 190, 222, 

324 

rolling on plane, 450-453, 454 
Displacement, of rigid body, 61, 62, 
70, 278-285, 290, 301, 302 
reduced to translation and 
rotation, 61, 62, 280, 281, 302 
screw, 284, 285 
virtual, 53, 58, 461 
Distributive property of vector 

operations, 20, 246, 249 
Disturbing force, 162, 163, 166-168, 

184, 186 
dn x, 368, 412 
Drag, 272 

Dugan, R. S., 13, 429 
Duncan, W. J., 316 
Dynamical unit of force, 34, 502 
Dynamics, plane, applications in, 

151-184-186, 189-222, 223 
methods of, 127-146, 147 
in relativity, 491-498, 499 
in space, applications in, 364-411- 

414, 418-453, 464 
methods of, 337-359, 360, 361 
(See also Motion; Particle; 
Rigid body; System of 
particles) 
Dyne, 34 



E 



Earth, angular velocity of, 143 
attraction of, 82, 84r-86, 144-146, 

404, 405 
models of, 5, 6, 84, 85, 144, 403, 

428, 429 
rotation of, 13, 143-146, 403-411, 

414 
Earth's axis, motion of, 428, 429 



Eccentric anomaly, 187 

Effective force, 138 

Einstein, A., 7, 475, 477, 478 

Elastic beam, 95-98, 114 

Elastic bodies in collision, 231-235, 

239 
Electric field, axially symmetric, 387, 

390-392, 396 
uniform, 383-386, 412 
Electric lens, focal length of, 403 

(See also Electrostatic lens) 
Electric potential, 381 
Electric vector, 381 
Electromagnetic field, 381-403, 412, 

413 

axially symmetric, 387-403, 413 
uniform, 383-387, 412 
Electromagnetic lens, 392-403, 413 
approximations for, 397-403 
focal length of, 403 
magnification of, 396, 413 
Electron, determination of e/m for, 

386, 387 

Electron optics (see Charged particle) 
Electrostatic attraction, 176 
Electrostatic field, 381, 383-386, 387, 
390-392, 396, 412 

(See also Electric field) 
Electrostatic lens, 398 
Ellipse, momental, 322, 323 
Ellipsoid, momental, 315-322, 333 
moments of inertia of, 323-325 
of Poinsot, 420, 421, 425, 463 
Ellipsoidal shell, moments of inertia 

of, 326 
Elliptic cylinder and plate, moments 

of inertia of, 324 

Elliptic functions, 364-370, 411, 412 
Elliptic harmonic motion, 169, 374 
Elliptic integral, 368 
Elliptical orbit, 169, 179-182, 186, 

374 

Emde, F., 370 
Energy, principle of, 129-131, 136, 

146, 147, 196, 201, 223, 231, 341, 

342, 346, 352, 356, 360, 361, 382, 

495-497, 499 



INDEX 



519 



Energy in relativity, 495-497, 499 
total, 130, 137, 341, 346 

(See also Kinetic energy; Poten- 
tial energy) 

Equation of the hodograph, 156, 184 

Equations of motion, impulsive, 

229-231, 238, 357, 358, 361, 

470-472 

of charged particle, 382, 383, 386, 

388, 389 

Lagrange's 458-472 
of particle, in cylindrical coordi- 
nates, 338 
in a plane, 127, 128, 146, 461, 

462 
relative to rotating earth, 403- 

406, 414 
relative to rotating frame, 142, 

348-350 

in relativity, 493-498, 499 
in space, 337-342, 360 
of rigid body, with fixed axis, 196, 

223 

with fixed point, 351, 352, 360 
in general, 355, 356, 361 
moving parallel to a plane, 202, 

223 
Equilibrium, of particle, 39, 40, 69, 

70, 262, 301 
of rigid body, movable parallel to 

a fixed plane, 62-64, 70 
in space, 273-278, 301 
stability of, 214-222, 223, 296 
of system of particles, 41-52, 261- 

266, 295-301 

Equimomental systems, 326 
Equipollent force systems, 47-52, 
70, 260, 261, 266-273, 293, 301 
Equivalence, of Galilean frames, 480, 

481 

mechanical, 9, 10 

Equivalent force systems, 47, 63, 64 
Equivalent simple pendulum, 200, 

223 

Erg, 54 

Euler-Bernouilli, law, 96, 114 
theory of beams, 95-98, 114 



Eulerian angles, 288-290 
angular velocity in terms of, 309, 

310 

Eiiler's equations of motion, 352, 360 
Euler's theorem, 279, 280, 301 
Event, 11, 478, 498 
Ewald, P. P., 88 
Extension, 96 
External forces, 41, 42 



Ferel's law, 408 

Ferry, E. S., 445 

Fictitious forces, 138, 139, 141-144, 
147,347-351,406 

Field, scalar or vector, 28 

Field of force, 66 

electromagnetic, 381-403, 412, 413 
electrostatic, 381, 383-387, 390- 

392, 396, 412 

gravitational, 82-86, 144, 405 
magnetostatic, 381, 382, 383-388, 

390-392, 397, 412 
uniform, 67 

Finite displacement of rigid body, 
278-282, 301, 302 

Flexible cable (see Cable) 

Flywheel, 196-198 

Focal length, of electromagnetic 
lens, 403 

Foot, 13 

Force, 15-17, 36 
absolute, 495 
applied, 58, 295 
central, 128, 168-186, 462 
centrifugal, 143-147, 349, 350, 406 
on charged particle, 176, 382 
Coriolis, 143, 144, 147, 349, 406 
effective, 138 

external and internal, 41, 42 
fictitious, 138, 139, 141-144, 147, 

347-351, 406 
field of, 66, 67 
of friction, 87, 88 
generalized, 293-301, 302, 462, 
464-467, 472 



520 



INDEX 



Force of gravity, 82-86, 144-146, 

404, 509 
impulsive, 228-238, 357-359, 361, 

470-472 

in relativity, 479, 495^199 
reversed effective, 138, 147 
shearing, 92-97, 114, 272, 273 
total, 259, 285, 301 
transmissibiiity of, 64 
unit of, 16, 34, 502 
Force system, general, 259-261 
invariants of, 269, 270, 285 
reduction of, 50-52, 70, 266-273, 

301 

Force systems, equipollent, 47-52, 
70, 260, 261, 266-273, 293, 301 
equivalent, 47, 63, 64 
Forced oscillations, 166-168, 186 
Forces, parallelogram of, 32, 33, 36 
polygon of, 40 
triangle of, 40 

which do no work, 54-57, 70 
Foucault's pendulum, 408-411, 414 
Foundations of mechanics, 3-35, 36 
Frame of reference, 11, 12, 35, 477 
astronomical, 31, 33, 141, 143 
Galilean, 478, 479 
moving, 139-146, 147, 346-351, 

361 

Newtonian, 32-34, 132, 134, 147 
reduced to rest, 141-143, 147, 347, 

349, 350 
in relative motion in relativity, 

480-491 
rotating, 142, 143, 147, 347-351, 

361 

Frames, 106-113, 116 
analytical and graphical methods, 

113 

critical forms, 113 
just-rigid and over-rigid, 106, 107 
simple and complex, 111, 112 
summary of methods, 115 
Frazer, R. A., 316 

Free particle, in Newtonian mechan- 
ics, 32 
in relativity, 489, 490, 498 



Free vector, 18, 19 
Freedom, degrees of, 207, 208, 287 
Frenet-Serret formulas, 264 
Frequencies, normal, 211-214, 223, 

468, 469 
Frequency of harmonic oscillator, 

162, 163 

(See also Periodic time) 
Friction, 86-92, 114 
angle of, 87, 88, 114 
coefficient of, 87, 88, 114 
cone of, 87 
limiting, 88 
Function denned by differential 

equations, 364 
Fundamental plane, 39 
Future and past, determination of, 
340, 341 

G 

0, 85, 86, 114, 144-146, 405 
Galilean frame of reference, 478, 479 
General theory of relativity (see 

Relativity) 
Generalized coordinates, 285-292, 

302, 462-472 
Generalized forces, 293-301, 302, 

462, 464-467, 472 
Generalized impulsive forces, 470- 

472 

Geodesic, 265, 266, 339 
Gradient vector, 28-31, 35, 36 
Gram, 13 

Gravitation in relativity, 477 
Gravitational attraction, 82-86, 114, 

144-146, 176-177, 404, 405, 509 
Gravitational constant (<?), 82-84, 

176, 509, 510 

Gravity, center of, 84-86, 114, 271 
Growth of vector, 348 
Gyration, radius of, 189 
Gyrocompass, 444-447, 454 
Gyroscope, 429, 441-447, 464 
Gyroscopic couple, 442-444, 464 
Gyroscopic effect of rotary engine, 

444 
Gyrostat (see Gyroscope) 



INDEX 



521 



H 

Hamilton, W. R., 33 
Harmonic oscillator, 159-168, 184, 
186 

with constant disturbing force, 
162, 184 

damped, 163-168, 186 

forced oscillations of, 166-168, 186 
Hemisphere, mass center of, 78, 81 
Hemispherical shell, mass center of, 

81,82 

Herpolhode cone, 309 
Heterogeneous body, 76 
Hodograph, 120, 121, 125, 156, 184 
Holonomic system, 287 
Homogeneous body, 76 
Hooke's joint, 297, 298 
Hooke's law, 96, 114 
Hoop, moment of inertia of, 190, 222 
Horizontal plane, 85 

or rotating earth, 145, 403 
Horsepower, 54 
Hyperbolic orbit, 179, 186 



Image, formed by electromagnetic 

lens, 391, 395-403, 413 
Image plane, 395, 413 
Image point, 395 
Impulse, 227, 357 
Impulsive couple, 229 
Impulsive force, 228-238, 357-359, 

361, 470-472 
Impulsive moment, 230, 231, 238, 

358, 361 
Impulsive motion, 227-238, 239, 

356-359, 361, 470-472 
Ince, E. L., 393 
Inclined plane, 202-204 
Indeterminate problems, 68, 69, 90, 

278 
Inertia, moments of (see Moments 

of inertia) 

products of (see Products of 
inertia) 



Infinitesimal displacement of rigid 
body, 61, 62, 70, 281-285, 290, 
302 
Ingredients, of mechanics, 8-17, 36 

of relativity, 476-478 
Instability (see Stability) 
Instantaneous axis, 308 
Instantaneous center, 123-126 
Integration of equations of motion 

in power series, 339, 340 
Intensity of wrench, 269 
Internal forces, 41, 42 
Internal reactions, in beam, 92, 93, 
114, 272, 273 

in flexible cable, 98 

in rigid body, 56, 57, 70, 206, 207 
Intrinsic equation of catenary, 102 
Intrinsic equations of motion, 338 
Invariable line, 419 
Invariable plane, 420, 463 
Invariant element in space-time, 489 
Invariants, of force system, 269, 270, 
285 

of infinitesimal displacement, 285 
Inverse square law, 176-186 
Lsotropy of Galilean frame, 478, 479 



Jacobian elliptic functions, 364-370, 

411, 412 
Jahnke, K., 370 
Joints, in a frame, 106, 107 
method of, 108-110, 116 
Just-rigid frame, 106, 107 

K 

Kaufmann, method of, 387 
Kelvin's theorem, 363 
Kepler's laws, 181, 182 
Kinematics, of particle, 118-121, 

126, 305-308, 332 
in relativity, 485-495, 498 
plane, 118-126 
of rigid body, 121-126, 308-313, 

332 
in space, 305-313, 332 



522 



INDEX 



Kinetic energy, of mass center, 195 
of particle, 129, 146, 333, 460 
of rigid body, 193-196, 222, 223, 

327-329, 333 
of system, 136, 137, 147, 463 

Konig, theorem of, 195 



Lagrange's equations, 458-472 
applications of, 467-472 
for general system, 466, 467, 472 
for impulsive motion, 470-472 
for particle in a plane, 459-462 
for system with two degrees of 
freedom, 463-466 

Lamb, H., 16, 113, 281, 445 

Lamina, representative, 61 

Lamy's theorem, 40 

Laplace's equation, 381, 382 

Latitude, astronomical, 145, 403 

Law, of action and reaction, 32, 36 
of motion, 32, 33, 36, 140-143, 

147, 347, 349, 350 
of the inverse square, 176-185 
of the parallelogram of forces, 32, 
33, 36 

Laws, of friction, 87, 88 

of Newtonian mechanics, 31-34, 
36 

Left-handed triad, 247 

Length, 11 

unit of, 11, 13, 14, 501, 502 

Lens, electric, 403 

electromagnetic, 392-403, 413 
magnetic, 403 

(See also Electrostatic lens; 
Magnetostatic lens) 

Level surface, 29 

Lift, 272, 512, 513 

Light, in relativity, 479-481, 483, 

485, 489, 496, 498 
speed of, 27, 483 

Limiting velocity, 159 

Line density, 77 

Linear moment, 74 



Linear momentum, in impulsive 
motion, 229, 230, 238, 357, 358, 
361 

of particle, 128, 337, 495, 496 
of system, 132-134, 146, 147, 343, 

360 

Linked rods, 236-238, 471, 472 
Loaded string, vibrations of, 208- 

212, 511, 512 

Logarithmic decrement, 166 
Lorentz transformation, 480-491, 

498 
Luneberg, R. K., 388 

M 

Mach, E., 16 

Magnetic field, axially symmetric, 

387, 388, 390-392, 397 
uniform, 383-387, 412 
Magnetic lens, focal length of, 403 
(See also Magnetostatic lens) 
Magnetic potential, 382 
Magnetic vector, 382 
Magnetostatic field, 381-388, 390- 

392, 397, 412 

Magnetostatic lens, 398, 413 
Mass, 9, 10 
of electron, 386, 387 
in relativity, 495 
unit of, 10, 13, 14, 501, 502 
Mass center, 74-82, 113, 114 
angular momentum relative to, 
135, 136, 147, 330, 345, 355, 
360, 361 
found by integration, 77, 81, 82, 

114 

found by symmetry and decom- 
position, 77-79, 114 
kinetic energy of, 195 
motion of, 132-134, 147, 230, 238, 

343, 360, 361 

motion relative to, 134-136, 147, 
234, 235, 238, 344, 345, 360, 
361 

of solar system, 134 
Mathematical models, 5, 6 



INDEX 



523 



Mathematical truth, 7 
Mathematical way of thinking, 4, 5, 

508 

Mathematicians* equations, 506, 507 
Matrix, 316 
Mean anomaly, 187 
Measuring rod or scale, 11, 477, 501, 

502 
relativistic contraction of, 486, 

487, 490 
Mechanical equivalence of bodies, 

9, 10 
Mechanics, foundations of, 3-35, 

36 

Mercury, 7, 31, 379 
Metacenter, 225 
Methods, of dynamics in space, 337- 

359, 360, 361 

of plane dynamics, 127-146, 147 
of plane statics, 38-69, 70 
Michelson, A. A., 481 
Michelsori-Morlcy experiment, 481 
Milne-Thomsoii, L. M., 370 
Minimum of potential energy, 214 

220, 223, 296 

Mixed triple product, 250, 251, 267 
Model, mathematical, 5-7 
used for prediction of full-scale 

phenomena, 512, 513 
Modes of vibration, normal, 207- 

214, 223 

Modulus, compression, 186 
of elliptic functions, 367 
Young's, 96, 114 
Moment, bending, 92-98, 114, 272, 

273 

of couple, 49, 50, 267 
impulsive, 230, 231, 238, 358, 361 
linear, 74 
of momentum, 128 

(See also Angular momentum) 
pitching, 272 
total, 259, 285, 301 
of vector, about line, 43-45, 255- 

257 

in plane mechanics, 44, 70 
about point, 252-255, 267 



Momental ellipse, 322, 323 
Momental ellipsoid, 315, 316, 318, 

321, 322, 333 
Moments of inertia, 189-193, 222, 

313-326, 333 

found by decomposition and differ- 
entiation, 325, 326 
principal, 316, 318-320, 322-325, 

333 
of simple bodies, 190-193, 222, 

323-325 

Momentum (sec Angular momen- 
tum ; Linear momentum) 
Morley, E. W., 481 
Motion, defined, 14 

of charged particle, 381-403, 412, 

413 

impulsive (see Impulsive motion) 
of mass center, 132-134, 147, 230, 

238, 343, 360, 361 
of particle, under central force, 

168-184, 186, 462 
determined by initial condi- 
tions, 339, 340 
in plane, 118-121, 126, 127-131, 

146, 151-184-186, 212-214, 
459-462 

relative to moving frame of 
reference*, 139-143, 147, 346- 
351 

in relativity, 489-498, 499 
in space, 305-308, 337-343, 360, 

364-411-414 
relative to mass center, 134-136, 

147, 234, 235, 238, 344, 345, 

360, 361 

of rigid body, parallel to fixed 
plane, 121-124, 126, 189-222, 
223 

with fixed point, 308-310, 332, 
351-355, 360, 418r-444, 463, 
454 
general, 310-313, 332, 355, 356, 

361, 447-453, 464 

of system, 131-139, 146, 147, 189- 

222, 223, 343-346, 360 
Moving constraints, 474 



524 



INDEX 



Moving frames of reference, 139- 

146, 147, 340-351, 361 
Moving rod, contraction of, 486, 487, 

490 
Multiplication, of vector and scalar, 

19,20 

of vectors, 245-267 
Murnaghan, F. D., 16 
Myers, L. M., 392 

N 

Necessary conditions of equilibrium, 

39, 40, 42, 43, 45-47, 59, 60, 69, 

70, 262, 263, 273-275, 296, 301 
Neutral equilibrium, 216 
Newton, L, 32, 82, 176, 475 
Newtonian frame of reference, 32- 

34, 132, 134, 147 
Newtonian law of gravitational 

attraction, 82 
Newtonian mechanics, laws of, 31- 

35,36 

Newtonian unit of time, 12 
w-gonal symmetry, 322 
Non-holonomic system, 287, 466 
Normal components of velocity and 

acceleration, 118, 119, 125, 305, 

306, 332 
Normal frequencies and periods, 

211-214, 223, 468, 469 
Normal modes of vibration, 207-214, 

223 

Normal reaction, 87 
Normal vector, principal, 264 
Notation, dimensional, 503 

for vectors, 19 
Null lines in space-time, 489 
Null planes and lines in statics, 304 
Nutation of top, 432 

O 

Object plane, 395 
Object point, 395 
Observer in relativity, 477 
Optics, electron (see Charged 
particle) 



Orbit, central, 166-184, 185, 186 

circular, 174-176 

elliptical, 169, 179-182, 186, 374 
(See also Planetary orbit) 

hyperbolic or parabolic, 179, 185 
Ordered triad, 247 
Orthogonal triad, 23 
Oscillation, center of, 201 
Oscillations, damped, 163-168, 186 

deadbeat, 166, 186 

forced, 166-168, 185 

harmonic, 160-162, 184 

(See also Pendulum; Vibration) 
Oscillator, harmonic, 159-168 
Osculating plane, 264 
Over-rigid frame, 106 



Pappus, theorems of, 80, 114 
Parabola in suspension bridge, 100, 

114 

Parabolic orbit, 179, 186 
Parabolic trajectory of projectile, 

151-154, 184 
Parallel axes, theorem of, 191-193, 

222 

Parallel forces, 270, 271 
Parallelepiped (see Cuboid) 
Parallelogram of forces, 32, 33, 36 
Parallelogram law, for couples, 268 

for infinitesimal rotations, 282 
Particle, 8, 9, 36 

angular momentum of, 128, 129, 

146, 329, 333, 341, 360 
under central force, 128, 168-184, 

185, 462 

charged, 176, 381-403, 412, 413 
dynamics of, 127-131, 146, 151 
184-186, 212-214, 337-343, 
360, 364-411-414, 459-462, 
491-198, 499 
equilibrium of, 39, 40, 69, 70, 262, 

301 

free, 32, 489, 490, 498 
kinematics of, 118-121, 125, SOS- 
SOS, 332, 485-495, 498 



INDEX 



525 



Particle, kinetic energy of, 129, 146, 

333, 460 

Lagrange's equations for, 459-462 
linear momentum of, 128, 337, 

495, 496 

in a plane, 64-66, 118-121, 126, 
127-131, 146, 151-184-186, 
212-214, 217, 218, 370-372, 
459-462 

potential energy of, 65-67 
principle of energy for, 129-131, 
146, 341, 342, 360, 382, 496, 
497, 499 

in relativity, 477, 489-498, 499 
on rotating earth, 144- J 46, 403- 

411, 414 
in rotating frame, 142, 143, 147, 

347-351, 361 

in space, 305-308, 329, 332, 333, 
337-343, 346-351, 360, 373- 
411-414 

on stretched stung, 511 
Particles, on stretched string, 208- 

212 

system of (see System of particles) 
Past and future, determination of, 

339, 340 

Pendulum, ballistic, 235, 236 
compound, 198-201, 223, 370 
conical, 374, 375 
equivalent simple, 200, 223 
Foucault's, 40&-411, 414 
simple, 159-161, 184, 370-372, 

412 

spherical, 373-381, 412 
Percussion, center of, 239 
Perihelion, 180 
Period (see Periodic time) 
Periodic solutions of a differential 

equation, 364-367 
Periodic time, 161 
of compound pendulum, 200, 201, 

223, 370 
of harmonic oscillator, 162, 163, 

166, 168 

normal, 211-213, 223, 468, 469 
of planet, 180-182, 186 



Periodic time of simple pendulum, 

161,184,372,412 
Perpendicular axes, theorem of, 193, 

222 

Phase of oscillator, 162 
Philosophical ideas, 3-8 
Physical laws and dimensions, 508- 

512 

Physical truth, 7 

Physical way of thinking, 3-5, 508 
Physicists' equations, 506, 507 
Pitch, of screw displacement, 284, 
285 

of wrench, 269, 270, 285 
Pitching moment, 272 
Plane, fundamental, 39 

inclined, 202-204 

invariable, 420, 463 

osculating, 264 

of symmetry, 78, 321 
Plane dynamics, applications in, 
151-184-186, 189-222, 223 

methods of, 127-146, 147 
Plane equipollence, 48 
Plane impulsive motion, 227- 238, 239 
Plane kinematics, 118-125 
Piano mechanics denned, 39 
Plane 4 statics, applications in, 74- 
113 116 

methods of, 38-69, 70 
Planetary orbit, 176-184, 186 

constants of, 179, 180, 186 

Kepler's laws for, 181, 182 

periodic time of, 180, 181, 186 
Plumb line oil lotating earth, 145, 

146, 403, 405 
Poinsot, method of, 419-421, 425- 

429, 463 

Poinsot ellipsoid, 420, 421, 425, 463 
Polhode cone, 309 
Polygon of forces, 40 
Poschl, Th., 88 
Position vector, 24, 36, 305 
Positive rotation, 247 
Potential, electric, 381 

gravitational, 83 

magnetic, 382 



526 



INDEX 



Potential energy, 64-67, 70, 85, 114, 
130, 294-297, 299, 300, 302 
for inverse square law of attrac- 
tion, 83, 178 
a minimum for stability, 214-221, 

223, 296 
Pound, 13 
Poundal, 34 
Power, 54 
Prandtl, L., 88 
Precession, 429-432, 438, 442-444, 

463, 454 
Principal axes and moments of 

inertia, 316-326, 333 
Principal normal, 264 
Principal planes of inertia, 318 
Principle of angular momentum, in 
impulsive motion, 230, 231, 238, 
357, 358, 361 

for particle, 128, 129, 146, 341, 360 
relative! to mass center, 136, 147, 

202, 223, 345, 360, 361 
for rigid body, 196, 202-204, 223, 

352, 355, 360, 361 
for system, 135, 147, 344, 345, 360 
Principle of energy, for particle, 
129-131, 146, 341, 342, 360, 382 
in relativity, 496, 497, 499 
for rigid body, 196, 201, 223, 352, 

356, 360, 361 

for system, 136, 137, 147, 346, 360 
Principle of equivalence, 480, 481 
Principle of linear momentum, in 
impulsive motion, 229, 230, 
238, 357, 358, 361 
for particle, 337, 496 
for system, 132, 146, 343, 360 
Principle of virtual work, 57-60, 70, 

295, 296, 301 
Procedure in theoretical mechanics, 

6,7 

Products of inertia, 313-316, 333 
Products of vectors, 245-267 
mixed triple, 250, 251, 267 
scalar, 245, 246, 267 
vector, 247-250, 267 
vector triple, 252, 267 



Projectile, with resistance, 154-159, 

184 

without resistance, 151-154, 184 
on rotating earth, 407, 408, 414 
stability of, 440, 441 

Propeller, 312, 313, 321, 322 

Proper energy, 497 

Proper mass, 495 

Proper time, 491, 492, 498 

Q 

Quantum mechanics, 7, 8, 177, 340, 
341 

R 

Radial components of velocity and 

acceleration, 120, 126 
Radius, of curvature, 119, 264 
of gyration, 189 
of torsion, 264 
Range of projectile, 153, 154 
Rate, of change of vector, 347, 348, 

361 

of growth, 348 
of transport, 348 
Rawlings, A. L., 445 
Reaction, in beam, 92-94, 114, 272, 

273 

normal, 87 

in rigid body, 56, 57, 70 
in rotating rod, 206, 207 
at rough contact, 86-88, 114 
at smooth contact, 54, 55, 57, 70 
workless, 54-57, 70 
Reactions of constraint, 57-59, 295 
Rectangular cuboid, moments of 

inertia of, 324 
Rectangular plate, moments of 

inertia of, 190, 222, 324 
Reduction, of displacement, to 

screw, 284, 285 
to translation and rotation, 61, 

62, 280, 281, 302 

of general force system, 266-273, 
301 



INDEX 



527 



Reduction, of plane force system, 

50-52, 70 
of system of parallel forces, 270, 

271 

Relative energy, 495, 497, 499 
Relative force, 495-499 
Relative mass, 495 
Relative momentum, 495, 496 
Relativistic contraction of moving 

rod, 486, 487, 490 
Relativistic slowing down of moving 

clock, 487, 490 
Relativity, fundamental concepts of, 

475-480 

general theory of, 7, 177, 476, 477 

measurement of time in, 475-480, 

483, 487, 490, 492, 493, 498 

motion of a particle in, 489-498, 

499 

special theory of, 475-498, 499 
Representative lamina, 61 
Resistance of air, 151, 154-159, 184 
varying as the square of the 

velocity, 157-159, 184 
Varying directly as the velocity, 

159, 184 

Resonance, 163, 168 
Rest, 14 

Rest energy, 497 
Restitution, 233, 234 

coefficient of, 232-234, 239 
Resultant, of finite rotations, 20, 282 
of forces, 32, 36 
of infinitesimal displacements, 

281-283, 302 

Reversed effective force, 138, 147 
Right-handed triad, 247 
Rigid body, 10, 11, 35 
angular momentum of, 193, 194, 

222, 223, 330-332, 333 
angular velocity of, 122, 126, SOS- 
SI 1, 332 
displacement of, 61, 62, 70, 278- 

285, 301, 302 

dynamics of, 189-207, 221-223, 
351-356, 360, 361, 418-463, 
464 



Rigid body, equilibrium of, 62-64, 

70, 273-278, 301 
free, 290 
internal reactions in, 56, 57, 70, 

206, 207 

kinematics of, 121-124, 126, SOS- 
SIS, 332 
kinetic energy of, 193-196, 222, 

223, 327-329, 333 
motion parallel to a plane, 121- 

124, 126, 189-207, 221-223 
motion in space, 351-356, 360, 

361, 418-463, 464 
in relativity, 477 
rotating about fixed axis, 196-201, 

223 
work done by forces acting on, 63, 

70, 292, 293, 302 

Rigid body with a fixed point, 

angular momentum of, 330-333 

angular velocity of, 308-310, 332 

displacement of, 279-282, 290, 

301, 302 

dynamics of, 351-355, 360, 418- 

444, 463, 464 

equations of motion of, 352, 360 
Kulerian angles for, 288-290, 309, 

310 

Eulor's theorem for, 279, 280, 301 
kinematics of, 308-310, 332 
kinetic energy of, 327, 328, 333 
mounting of, 418 
under no forces, 418-429, 463 

(See also Gyroscope; Top) 
Rod, moment of inertia of, 190, 222-. 
Rolling contact, 55-57, 70, 123, 124 
Rolling disk, 450-453, 464 
Rotating frame of reference, 142, 

143, 147, 347-351, 361 
Rotating rod, 206, 207 
Rotation, of the earth, 13, 143-146, 

403-411, 414 

about fixed axis, 196-201, 223 
about fixed point, 279-282, 301, 

302, 308-310, 332 
instantaneous axis of, 308 
in a plane, 61, 62, 121-126 



528 



INDEX 



Ilotation, positive, 247 
Rotations, finite, resultant of, 20, 282 
infinitesimal, resultant of, 281, 

282, 302 

Rough contact, 85-88, 114 
Routh's rule, 325 
Russell, H. N., 13, 429 

S 

Scalar, 18 

multiplied by a vector, 19, 20 
Scalar field, 28 
Scalar product, 245, 246, 249, 250, 

257 

Screw displacement, 284, 285 
Second, 13, 14 

Sections, method of, 110, 111, 116 
Semicircular plate and wire, mass 

centers of, 80 

Separation in space-time, 489, 498 
Shearing force, 92-97, 114, 272, 273 
Shell (see Projectile) 
Significant figures, 508 
Silberstein, L., 481 
Simple frame, 111 
Simple harmonic motion, 160-162, 

184 

Simple pendulum, equivalent, 200, 
223 

finite oscillations of, 161, 200, 370- 
372, 412 

small oscillations of, 159-161, 184 
Sleeping top, 438-440, 463 
Sliding vector, 18 
Slowing down of moving clock, 487, 

490 

Small displacement (see Infinites- 
imal displacement) 
Smooth contact, 54, 55, 57, 70 
sn x, 367-370, 411, 412 
Solar system, dynamics of, 182 

mass center of, 134 
Sommerville, D. M. Y., 320 
Space centrode, 124, 126, 309 
Space cone, 309, 421, 425, 427, 463 
Space-time, 487-491, 498 

vectors in, 494, 495 



Special theory of relativity (see 

Relativity) 
Speed, 27 

of approach, 232, 233, 239 
of light, 27, 479-481, 483, 485, 496 
of separation, 232, 233, 239 
Sphere, mass center of, 78 

moment of inertia of, 191, 222, 324 
Spheres, collision of, 231-235 
Spherical pendulum, 373-381, 412 
apse of, 378-381 
general motion of, 375-378, 412 
small oscillations of, 373, 374, 378- 

381,412 

Spherical polar coordinates, 467, 468 
Spherical shell, attraction of, 83, 84 

moment of inertia of, 326 
Spin of top or gyroscope, 430, 433, 

442-444, 463, 464 
Spinning top (see Top) 
Stability, of circular orbit, 174-176 
of equilibrium, 214-221, 223, 296 
of gyroscope, 441, 442, 464 
of rolling disk, 450-453, 464 
of sleeping top, 43&-440, 453 
of spinning projectile, 440, 441 
Statically determinate problems for 

beams, 94, 95 
Statically indeterminate problems, 

68, 69, 90, 278 
Statics, plane, applications in, 74- 

113-116 

methods of, 38-69, 70 
in space, 259-301 ; 302 
Stewart, ,1. Q., 13, 429 
Stress, in bar of frame, 108 

in beam, 92, 93, 114, 272, 273 
String (sec Cable) 
Subtraction of vectors, 21 
Sufficient conditions of equilibrium, 
39, 40, 59, 60, 62, 63, 69, 70, 
273-275, 301 
Surface, level, 29 
rough, 86-88, 114 
smooth, 54, 55, 57, 70 
Surface density, 77 
Suspension bridge, 100, 114 



INDEX 



529 



Symmetry, axis of, 78, 321, 322 
central, 77 

of central orbit, 172, 173, 185 
diagonal, trigonal, etc., 321 
plane of, 78, 321, 322 
used to find mass centers, 77, 78, 

114 
used to find principal axes, 320- 

323 

Synchronization of clocks, 477-480 
System of forces (see Force system) 
System of particles, angular momen- 
tum of, 134-136, 147, 329, 330, 
344, 345, 360 

dynamics of, 131-139, 146, 147, 

189-222, 223, 343-346, 360 

equilibrium of, 41-52, 57-67, 70, 

261-266, 295-301 
kinetic energy of, 136, 137, 147, 

463 

Lagrange's equations for, 463-472 
linear momentum of, 132-134, 

146, 147, 343, 360 
potential energy of, 64-67, 137, 
294-297, 299, 300, 302 

T 

Tangential components of velocity 
and acceleration, 118, 119, 125, 
305, 306, 332 
Tension, in bar of frame, 108 

in beam, 92-96, 114, 272 

in cable, 98-105, 114, 116, 265, 266 
Tensor, 316 

Tetragonal symmetry, 321 
Tetrahedron, mass center of, 79 
Theory of dimensions, 501-513 
Theory of relativity (see Relativity) 
Thin beams, 92-98, 114 
Thrust, 108 

Time, in Newtonian mechanics, 12, 
13 

proper, 491, 492, 498 

in relativity, 476-480, 483, 487, 
490-493, 498 

unit of, 12-14, 501, 502 
rimoshenko, S., 113 



Top, 429-441, 453 

cuspidal motion of, 436-438 

general motion of, 432-436, 463, 
469, 470 

Lagrange's equations for, 469, 470 

sleeping, 438-440, 463 

in steady precession, 429-432, 453 
Torque, 196 
Torsion, radius of, 264 
Total angular impulse, 357 
Total energy, 130, 137, 341, 346 
Total force, 259, 285, 301 
Total impulse, 357 
Total impulsive force, 357 
Total moment, 259, 285, 301 
Trajectory, of charged particle, 384, 
386, 389-392, 412, 413 

of projectile, 151-159, 184, 407, 

408, 414 

Transformation, of axes in space 
time, 488 

Lorentz, 480-491, 498 

Newtonian, 486 

to principal axes of inertia, 318- 

320, 322, 323 
Tninslation, 61, 279 
Transmissibili ty of force, 64 
Transport, acceleration of, 349 

of vector, 348 
Transverse* components of velocity 

and acceleration, 120, 125 
Triad, left-handed and right-handed, 
247 

ordered, 247 

unit orthogonal, 23 
Triangle of forces, 40 
Triangular plate, mass center of, 79 
Trigonal symmetry, 321 
Triple products, 250-252, 257 
Trusses (see Frames) 
Truth, mathematical and physical, 7 
Twisting couple, 272 
Two-body problem, 182-184, 186 

U 

Uniform field of force, 67 
electromagnetic, 383-387, 412 



530 



INDEX 



Unit coordinate vectors, 23, 247 
Unit, of force, 16, 34, 36, 502 
of length, 11, 13, 14, 35, 501, 502 
of mass, 10, 13, 14, 36, 501, 502 
of time, 12-14, 36, 501, 502 
Unit orthogonal triad, 23 
Units, arbitrariness of, 36, 502 
c.g.s. and f .p.s., 13 
change of, 504-508 



Varignon, theorem of, 44, 45, 255, 

26 
Vector, 17-19 

binormal, 264 

bound, 18, 19 

components of, 22-24, 32, 36, 36, 
245 

differentiation of 24-26, 36, 250 

electric, 381 

free, 18, 19 

gradient, 28-31, 36, 36 

magnetic, 382 

moment of, 43-45, 70, 252-267 

multiplied by scalar, 19, 20 

notation for, 19 

position, 24, 36, 305 

principal normal, 264 

rate of change of, 347, 348, 361 

sliding, 18 

in space-time, 494, 495 

zero, 21 

Vector field, 28 
Vector function, 24-26 
Vector product, 247-250, 267 
Vector triple product, 252, 267 
Vectors, addition of, 19-22, 36 

coordinate, 23, 247 

products of, 245-267 

subtraction of, 21 
Vehicle, self-propelled, 204^206 
Velocities, composition of, 140, 307, 

308, 491 
Velocity, 26-28, 36, 305 

absolute, 494 

angular, 122, 126, 308-311, 332 



Velocity, areal, 170, 180, 188 
in cylindrical coordinates, 306, 307 
of light, 27, 479-481, 483, 485, 496 
limiting, 159 
of particle of rigid body, 308-313, 

332 
radial and transverse components 

of, 120, 126 
tangential component of, 118, 126, 

305, 306, 332 
Vertical, 85 

on rotating earth, 145, 403 
Vibration, normal modes of, 207- 

214, 223 

of particle in plane, 212-214 
of particle on stretched string, 511 
of two particles on stretched 

string, 208-212 
(See also Oscillations) 
Virtual displacement, 53, 58, 60, 461 
Virtual work, 57-60, 70, 111, 116, 
295, 296, 301 

W 

Ways of thinking, 3-5, 508 
Weight, 17, 86 

on rotating earth, 145, 404, 405 
Whittaker, E. T., xi, 16 
Work, 53-67, 70, 292-301, 302 

done by couple, 64, 293 

done by force, 53, 70 

done by forces on general system, 
293, 294, 302 

done by forces on rigid body, 62, 

63, 70, 292, 293, 302 
Work, virtual, 57-60, 70, 111, 116,. 

295, 296, 301 

Workless constraints, 54-. r ) 
Wrench, 269, 270, 285, 30i 



Young, D. H., 113 
Young's modulus, 96, 114 



Zero, force equipollent to, 48, 261 
Zero vector, 21